OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral 3 Exercise 15 (A)

S Chand Class 12 ICSE Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(A)

Question 1.
(i) \( \int \frac{1}{x^2+36} \, dx \)
(ii) \( \int \frac{dx}{1+\frac{x^2}{4}} \)
(iii) \( \int \frac{dx}{50+2 x^2} \)
Answer:
(i) To solve this integral, we first rewrite the denominator to fit the standard form \( x^2+a^2 \).
\( \int \frac{1}{x^2+36} \, dx = \int \frac{1}{x^2+6^2} \, dx \)
Then, we apply the standard integration formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1} \frac{x}{a} + C \), with \( a = 6 \).
\( = \frac{1}{6}\tan^{-1} \frac{x}{6} + C \)
(ii) For the second integral, we begin by simplifying the denominator to align it with the \( a^2+x^2 \) form.
\( \int \frac{dx}{1+\frac{x^2}{4}} = \int \frac{4}{4+x^2} \, dx = \int \frac{4}{2^2+x^2} \, dx \)
Using the same standard formula with \( a=2 \), we get:
\( = \frac{4}{2}\tan^{-1}\frac{x}{2} + C = 2 \tan^{-1}\frac{x}{2} + C \)
(iii) In the third integral, we factor out the constant from the denominator to get it into the desired form \( a^2+x^2 \).
\( \int \frac{dx}{50+2 x^2} = \int \frac{dx}{2(25+x^2)} = \frac{1}{2}\int \frac{dx}{5^2+x^2} \, dx \)
Applying the standard formula with \( a=5 \), the result is:
\( = \frac{1}{2} \times \frac{1}{5} \tan^{-1} \left(\frac{x}{5}\right) + C = \frac{1}{10} \tan^{-1} \left(\frac{x}{5}\right) + C \)
In simple words: All these problems were about transforming the integral into a recognizable form like \( \frac{1}{x^2+a^2} \) or \( \frac{1}{a^2+x^2} \) so that we could use a specific formula involving the arctangent function. The key steps were algebraic manipulation and then applying the correct formula.

🎯 Exam Tip: Always look for ways to manipulate the integrand algebraically to match one of the standard integral forms. Factoring out constants or clearing fractions in the denominator are common techniques. Do not forget to add the constant of integration, C, for indefinite integrals.

Question 2.
(i) \( \int \frac{dx}{x^2-4} \)
(ii) \( \int \frac{d x}{9 x^2-16} \)
(iii) \( \int \frac{d y}{25-16 y^2} \)
(iv) \( \int \frac{d y}{18-2 x^2} \)
Answer:
(i) For the first integral, we express the denominator as a difference of squares and apply the relevant standard formula.
\( \int \frac{dx}{x^2-4} = \int \frac{dx}{x^2-2^2} \)
Using the standard integral formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \), with \( a = 2 \):
\( = \frac{1}{2 \times 2} \log \left| \frac{x-2}{x+2} \right| + C \)
\( = \frac{1}{4} \log \left| \frac{x-2}{x+2} \right| + C \)
(ii) For the second integral, we first factor out the constant 9 from the denominator to get it into the form \( x^2-a^2 \).
\( \int \frac{d x}{9 x^2-16} = \int \frac{1}{9} \frac{d x}{x^2-\frac{16}{9}} = \frac{1}{9} \int \frac{d x}{x^2-\left(\frac{4}{3}\right)^2} \)
Applying the same formula \( \int \frac{dx}{x^2-a^2} \) with \( a = \frac{4}{3} \):
\( = \frac{1}{9} \times \frac{1}{2 \times \frac{4}{3}} \log \left| \frac{x-\frac{4}{3}}{x+\frac{4}{3}} \right| + C \)
\( = \frac{1}{9} \times \frac{3}{8} \log \left| \frac{3x-4}{3x+4} \right| + C \)
\( = \frac{1}{24} \log \left| \frac{3x-4}{3x+4} \right| + C \)
(iii) For the third integral, we factor out 16 from the denominator and rewrite the expression to match the form \( a^2-y^2 \).
\( \int \frac{d y}{25-16 y^2} = \frac{1}{16} \int \frac{d y}{\frac{25}{16}-y^2} = \frac{1}{16} \int \frac{d y}{\left(\frac{5}{4}\right)^2-y^2} \)
Using the standard integral formula \( \int \frac{dy}{a^2-y^2} = \frac{1}{2a} \log \left| \frac{a+y}{a-y} \right| + C \), with \( a = \frac{5}{4} \):
\( = \frac{1}{16} \times \frac{1}{2 \times \frac{5}{4}} \log \left| \frac{\frac{5}{4}+y}{\frac{5}{4}-y} \right| + C \)
\( = \frac{1}{16} \times \frac{2}{5} \log \left| \frac{5+4y}{5-4y} \right| + C = \frac{1}{40} \log \left| \frac{5+4y}{5-4y} \right| + C \)
(iv) For the fourth integral, we factor out 2 from the denominator and rewrite it to fit the form \( a^2-x^2 \).
\( \int \frac{d y}{18-2 x^2} = \frac{1}{2} \int \frac{d x}{9-x^2} = \frac{1}{2} \int \frac{d x}{3^2-x^2} \)
Using the standard integral formula \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \), with \( a = 3 \):
\( = \frac{1}{2} \times \frac{1}{2 \times 3} \log \left| \frac{3+x}{3-x} \right| + C \)
\( = \frac{1}{12} \log \left| \frac{3+x}{3-x} \right| + C \)
In simple words: These problems involved integrating rational functions where the denominator was a quadratic expression. By factoring and rewriting the denominators into forms like \( x^2-a^2 \) or \( a^2-x^2 \), we could use specific logarithmic integral formulas to find the solutions. Remember to simplify the constant factors and include the absolute value for the logarithm argument.

🎯 Exam Tip: Identify the correct standard integral form based on the denominator (difference of squares, sum of squares, etc.). Pay close attention to factoring constants and correctly identifying 'a' for each formula. Ensure you use absolute values for the argument of the logarithm function.

Question 3.
(i) \( \int \frac{dx}{(x+2)^2+1} \)
(ii) \( \int \frac{d x}{1+2(x+2)^2} \)
Answer:
(i) For the first integral, we use a simple substitution to transform it into a standard arctangent form. Let \( t = x+2 \), which means \( dt = dx \).
\( I = \int \frac{d t}{t^2+1^2} \)
Using the formula \( \int \frac{dt}{t^2+a^2} = \frac{1}{a}\tan^{-1} \frac{t}{a} + C \) with \( a=1 \):
\( = \tan^{-1} \left(\frac{t}{1}\right) + C \)
Substitute back \( t = x+2 \):
\( = \tan^{-1}(x+2) + C \)
(ii) For the second integral, we also use the substitution \( t = x+2 \), so \( dt = dx \). Then, we manipulate the denominator to fit the \( a^2+t^2 \) form.
\( I = \int \frac{d t}{1+2t^2} \)
Factor out 2 from the denominator:
\( = \frac{1}{2}\int \frac{d t}{\frac{1}{2}+t^2} = \frac{1}{2}\int \frac{d t}{\left(\frac{1}{\sqrt{2}}\right)^2+t^2} \)
Now apply the arctangent formula with \( a=\frac{1}{\sqrt{2}} \):
\( = \frac{1}{2} \times \frac{1}{\frac{1}{\sqrt{2}}} \tan^{-1} \frac{t}{\frac{1}{\sqrt{2}}} + C \)
\( = \frac{1}{2} \times \sqrt{2} \tan^{-1} (\sqrt{2}t) + C = \frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2}t) + C \)
Substitute back \( t = x+2 \):
\( = \frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2}(x+2)) + C \)
In simple words: These integrals required making a substitution, replacing part of the expression with a new variable. After substitution, we algebraically adjusted the denominators to match known integral formulas, mostly involving the arctangent function. Finally, we put the original variable back into the answer.

🎯 Exam Tip: Look for opportunities to simplify integrals using substitution, especially when you see a term like \( (ax+b)^n \) or \( (ax+b)^2 \). After substitution, ensure the resulting integral is in a standard form. Remember to substitute back to the original variable at the end.

Question 4.
(i) \( \int \frac{3 x^2}{x^6+1} \, dx \)
(ii) \( \int \frac{x^2}{1+x^4} \, dx \)
(iii) \( \int \frac{\cos x}{1+\sin ^2 x} \, dx \)
(iv) \( \int \frac{e^x}{1+e^{2 x}} \, dx \)
(v) \( \int \frac{d x}{e^x+e^{-x}} \)
(vi) \( \int \frac{e^{-x}}{16+9 e^{-2 x}} \, dx \)
(vii) \( \int \sqrt{e^x-1} \, dx \)
Answer:
(i) To solve this integral, we can rewrite the denominator as \( (x^3)^2+1 \) and then use a substitution. Let \( t = x^3 \), so \( dt = 3x^2 \, dx \).
\( \int \frac{3 x^2 \, dx}{x^6+1} = \int \frac{3 x^2 \, dx}{(x^3)^2+1} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{dt}{t^2+1^2} \)
Using the formula for \( \int \frac{dt}{t^2+a^2} \), we get:
\( = \tan^{-1} t + C \)
Substitute back \( t = x^3 \):
\( = \tan^{-1} (x^3) + C \)
(ii) For this integral, the provided solution uses the substitution \( t = x^2 \), which implies \( dt = 2x \, dx \). Thus, it solves for an integral with \( x \, dx \) in the numerator.
\( \int \frac{x \, dx}{1+x^4} = \int \frac{x \, dx}{1+(x^2)^2} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{dt}{2(1+t^2)} = \frac{1}{2}\int \frac{dt}{1+t^2} \)
Applying the arctangent formula:
\( = \frac{1}{2}\tan^{-1} t + C \)
Substitute back \( t = x^2 \):
\( = \frac{1}{2}\tan^{-1} (x^2) + C \)
(iii) We use the substitution \( t = \sin x \), which means \( dt = \cos x \, dx \).
\( \int \frac{\cos x \, dx}{1+\sin ^2 x} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{dt}{1+t^2} \)
Applying the arctangent formula:
\( = \tan^{-1} t + C \)
Substitute back \( t = \sin x \):
\( = \tan^{-1} (\sin x) + C \)
(iv) Here, we let \( t = e^x \), which gives \( dt = e^x \, dx \). The term \( e^{2x} \) becomes \( (e^x)^2 = t^2 \).
\( \int \frac{e^x \, dx}{1+e^{2 x}} = \int \frac{e^x \, dx}{1+(e^x)^2} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{dt}{1+t^2} \)
Applying the arctangent formula:
\( = \tan^{-1} t + C \)
Substitute back \( t = e^x \):
\( = \tan^{-1} (e^x) + C \)
(v) First, rewrite the integral by multiplying the numerator and denominator by \( e^x \). Let \( t = e^x \), so \( dt = e^x \, dx \).
\( \int \frac{d x}{e^x+e^{-x}} = \int \frac{e^x \, dx}{e^{2x}+1} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{dt}{t^2+1} \)
Applying the arctangent formula:
\( = \tan^{-1} t + C \)
Substitute back \( t = e^x \):
\( = \tan^{-1} (e^x) + C \)
(vi) We use the substitution \( t = e^{-x} \), which gives \( dt = -e^{-x} \, dx \). Thus, \( -dt = e^{-x} \, dx \). Also, \( e^{-2x} = (e^{-x})^2 = t^2 \).
\( \int \frac{e^{-x} \, dx}{16+9 e^{-2 x}} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{-dt}{16+9t^2} = -\frac{1}{9}\int \frac{dt}{\frac{16}{9}+t^2} \)
Rewrite the denominator to match \( a^2+t^2 \):
\( = -\frac{1}{9}\int \frac{dt}{\left(\frac{4}{3}\right)^2+t^2} \)
Apply the arctangent formula with \( a=\frac{4}{3} \):
\( = -\frac{1}{9} \times \frac{1}{\frac{4}{3}} \tan^{-1} \left(\frac{t}{\frac{4}{3}}\right) + C \)
\( = -\frac{1}{9} \times \frac{3}{4} \tan^{-1} \left(\frac{3t}{4}\right) + C = -\frac{1}{12} \tan^{-1} \left(\frac{3t}{4}\right) + C \)
Substitute back \( t = e^{-x} \):
\( = -\frac{1}{12} \tan^{-1} \left(\frac{3e^{-x}}{4}\right) + C \)
(vii) This integral requires a more complex substitution. Let \( t = \sqrt{e^x-1} \). Squaring both sides gives \( t^2 = e^x-1 \).
From this, \( e^x = t^2+1 \). Differentiating both sides with respect to \( x \), \( e^x \, dx = 2t \, dt \).
So, \( dx = \frac{2t \, dt}{e^x} = \frac{2t \, dt}{t^2+1} \).
\( \int \sqrt{e^x-1} \, dx = \int t \times \frac{2t \, dt}{t^2+1} = \int \frac{2t^2 \, dt}{t^2+1} \)
Now, we can perform polynomial division or rewrite the numerator:
\( = \int \frac{2(t^2+1)-2}{t^2+1} \, dt = \int \left(2 - \frac{2}{t^2+1}\right) \, dt \)
Integrate term by term:
\( = 2t - 2\tan^{-1} t + C \)
Substitute back \( t = \sqrt{e^x-1} \):
\( = 2\sqrt{e^x-1} - 2\tan^{-1} (\sqrt{e^x-1}) + C \)
In simple words: These problems show how various expressions can be integrated using substitution techniques. We choose a part of the expression to be 't', find its derivative, and then rewrite the whole integral in terms of 't'. After integrating, we substitute the original variable back. Some problems also required algebraic manipulation to fit standard integral forms.

🎯 Exam Tip: Mastering substitution is crucial for integration. Practice identifying appropriate substitutions, especially for expressions involving \( e^x \), trigonometric functions, or terms raised to powers. Remember to change the \( dx \) term correctly when substituting. For complicated integrands, sometimes a second substitution or algebraic manipulation like adding and subtracting terms might be needed.

Question 5.
(i) \( \int \frac{x^2-1}{x^2+4} \, dx \)
(ii) \( \int \frac{x^4}{x^2+1} \, dx \)
(iii) \( \int \frac{3 x^5}{1+x^{12}} \, dx \)
(iv) \( \int \frac{d x}{2+\cos x} \, dx \)
Answer:
(i) To solve this, we can split the fraction by adding and subtracting 4 in the numerator.
\( \int \frac{x^2-1}{x^2+4} \, dx = \int \frac{x^2+4-5}{x^2+4} \, dx = \int \left(1 - \frac{5}{x^2+4}\right) \, dx \)
Separate the terms and integrate:
\( = \int 1 \, dx - 5 \int \frac{dx}{x^2+2^2} \)
Using the formula for \( \int 1 \, dx \) and \( \int \frac{dx}{x^2+a^2} \):
\( = x - 5 \times \frac{1}{2}\tan^{-1} \frac{x}{2} + C \)
\( = x - \frac{5}{2}\tan^{-1} \frac{x}{2} + C \)
(ii) For this integral, we use a similar technique of adding and subtracting 1 in the numerator to simplify the fraction.
\( \int \frac{x^4}{x^2+1} \, dx = \int \frac{x^4-1+1}{x^2+1} \, dx \)
Split the fraction and factor the numerator \( (x^4-1) \):
\( = \int \frac{x^4-1}{x^2+1} \, dx + \int \frac{1}{x^2+1} \, dx = \int \frac{(x^2-1)(x^2+1)}{x^2+1} \, dx + \int \frac{1}{x^2+1} \, dx \)
Simplify and integrate:
\( = \int (x^2-1) \, dx + \int \frac{1}{x^2+1^2} \, dx \)
\( = \frac{x^3}{3} - x + \tan^{-1} x + C \)
(iii) We use a substitution to simplify this integral. Let \( t = x^6 \), which implies \( dt = 6x^5 \, dx \). So, \( 3x^5 \, dx = \frac{1}{2} dt \).
\( \int \frac{3x^5}{1+x^{12}} \, dx = \int \frac{3x^5 \, dx}{1+(x^6)^2} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{\frac{1}{2} dt}{1+t^2} = \frac{1}{2}\int \frac{dt}{1+t^2} \)
Applying the arctangent formula:
\( = \frac{1}{2}\tan^{-1} t + C \)
Substitute back \( t = x^6 \):
\( = \frac{1}{2}\tan^{-1} (x^6) + C \)
(iv) This integral requires a specific substitution for trigonometric functions. Let \( t = \tan \frac{x}{2} \). Then \( dx = \frac{2 dt}{1+t^2} \) and \( \cos x = \frac{1-t^2}{1+t^2} \).
\( \int \frac{d x}{2+\cos x} = \int \frac{\frac{2 dt}{1+t^2}}{2+\frac{1-t^2}{1+t^2}} \)
Simplify the denominator:
\( = \int \frac{\frac{2 dt}{1+t^2}}{\frac{2(1+t^2)+1-t^2}{1+t^2}} = \int \frac{2 dt}{2+2t^2+1-t^2} = \int \frac{2 dt}{t^2+3} \)
Rewrite the denominator to match \( t^2+a^2 \):
\( = 2\int \frac{dt}{t^2+(\sqrt{3})^2} \)
Apply the arctangent formula with \( a=\sqrt{3} \):
\( = 2 \times \frac{1}{\sqrt{3}} \tan^{-1} \frac{t}{\sqrt{3}} + C \)
Substitute back \( t = \tan \frac{x}{2} \):
\( = \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{\tan \frac{x}{2}}{\sqrt{3}}\right) + C \)
In simple words: For these problems, we used various techniques like algebraic manipulation to simplify fractions (by adding and subtracting terms), substitution to change variables, and specific trigonometric substitutions for integrals involving \( \cos x \). The goal was always to transform the integral into a standard form that could be solved using known formulas.

🎯 Exam Tip: When the degree of the numerator is greater than or equal to the degree of the denominator, always perform polynomial long division or algebraic manipulation (like adding and subtracting terms) first. For integrals involving \( \sin x \) or \( \cos x \), the substitution \( t = \tan \frac{x}{2} \) is a universal method.

Question 6.
(i) \( \int \frac{x \, dx}{x^4-a^4} \)
(ii) \( \int \frac{x^2 \, dx}{a^6-x^6} \)
(iii) \( \int \frac{x}{1-x^4} \, dx \)
Answer:
(i) For the first integral, we use the substitution \( t = x^2 \), so \( dt = 2x \, dx \), which means \( x \, dx = \frac{1}{2} dt \).
\( \int \frac{x \, dx}{x^4-a^4} = \int \frac{x \, dx}{(x^2)^2-(a^2)^2} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{\frac{1}{2} dt}{t^2-(a^2)^2} = \frac{1}{2}\int \frac{dt}{t^2-(a^2)^2} \)
Using the standard formula \( \int \frac{dx}{x^2-b^2} = \frac{1}{2b}\log \left| \frac{x-b}{x+b} \right| + C \) with \( b=a^2 \):
\( = \frac{1}{2} \times \frac{1}{2a^2}\log \left| \frac{t-a^2}{t+a^2} \right| + C \)
Substitute back \( t = x^2 \):
\( = \frac{1}{4a^2}\log \left| \frac{x^2-a^2}{x^2+a^2} \right| + C \)
(ii) For the second integral, we let \( t = x^3 \), so \( dt = 3x^2 \, dx \), meaning \( x^2 \, dx = \frac{1}{3} dt \).
\( \int \frac{x^2 \, dx}{a^6-x^6} = \int \frac{x^2 \, dx}{(a^3)^2-(x^3)^2} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{\frac{1}{3} dt}{(a^3)^2-t^2} = \frac{1}{3}\int \frac{dt}{(a^3)^2-t^2} \)
Using the standard formula \( \int \frac{dx}{b^2-x^2} = \frac{1}{2b}\log \left| \frac{b+x}{b-x} \right| + C \) with \( b=a^3 \):
\( = \frac{1}{3} \times \frac{1}{2a^3}\log \left| \frac{a^3+t}{a^3-t} \right| + C \)
Substitute back \( t = x^3 \):
\( = \frac{1}{6a^3}\log \left| \frac{a^3+x^3}{a^3-x^3} \right| + C \)
(iii) For the third integral, we use the substitution \( t = x^2 \), so \( dt = 2x \, dx \), which means \( x \, dx = \frac{1}{2} dt \).
\( \int \frac{x}{1-x^4} \, dx = \int \frac{x \, dx}{1-(x^2)^2} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{\frac{1}{2} dt}{1-t^2} = \frac{1}{2}\int \frac{dt}{1^2-t^2} \)
Using the standard formula \( \int \frac{dx}{b^2-x^2} = \frac{1}{2b}\log \left| \frac{b+x}{b-x} \right| + C \) with \( b=1 \):
\( = \frac{1}{2} \times \frac{1}{2 \times 1}\log \left| \frac{1+t}{1-t} \right| + C \)
Substitute back \( t = x^2 \):
\( = \frac{1}{4}\log \left| \frac{1+x^2}{1-x^2} \right| + C \)
In simple words: These problems all involved using substitution to change the variable of integration, simplifying the integral into a standard form involving the difference of squares in the denominator. This allowed us to apply logarithmic formulas to solve them, always remembering to substitute the original variable back at the end.

🎯 Exam Tip: Integrals with \( x^4 \), \( x^6 \) or similar higher powers in the denominator often suggest a substitution like \( t=x^2 \) or \( t=x^3 \) to reduce the degree and match standard forms. Always check if the numerator provides the derivative (or a multiple of it) of your chosen substitution variable.

Question 7.
(i) \( \int \frac{x^3+x}{x^4-9} \, dx \)
(ii) \( \int \frac{\cos x}{4-\sin ^2 9} \, dx \)
(iii) \( \int \log \left(2+x^2\right) \, dx \)
(iv) \( \int \frac{x^2-4}{x^4+16} \, dx \)
Answer:
(i) We split the integral into two parts: one with \( x^3 \) and one with \( x \) in the numerator. Then we use substitution for each.
\( \int \frac{x^3+x}{x^4-9} \, dx = \int \frac{x^3}{x^4-9} \, dx+\int \frac{x \, dx}{x^4-9} \)
For the first part, let \( u = x^4-9 \), so \( du = 4x^3 \, dx \).
\( \int \frac{x^3}{x^4-9} \, dx = \frac{1}{4} \int \frac{4x^3 \, dx}{x^4-9} = \frac{1}{4} \int \frac{du}{u} = \frac{1}{4} \log|u| = \frac{1}{4} \log|x^4-9| \)
For the second part, let \( t = x^2 \), so \( dt = 2x \, dx \). The integral becomes \( \int \frac{\frac{1}{2} dt}{t^2-3^2} \).
\( \int \frac{x \, dx}{x^4-9} = \int \frac{x \, dx}{(x^2)^2-3^2} = \frac{1}{2}\int \frac{dt}{t^2-3^2} \)
Using the formula for \( \int \frac{dx}{x^2-a^2} \):
\( = \frac{1}{2} \times \frac{1}{2 \times 3}\log \left| \frac{t-3}{t+3} \right| = \frac{1}{12}\log \left| \frac{x^2-3}{x^2+3} \right| \)
Combining both parts:
\( I = \frac{1}{4} \log|x^4-9| + \frac{1}{12}\log \left| \frac{x^2-3}{x^2+3} \right| + C \)
(ii) We use the substitution \( t = \sin x \), which implies \( dt = \cos x \, dx \). The denominator is rewritten as \( 2^2-t^2 \).
\( \int \frac{\cos x \, dx}{4-\sin ^2 x} \)
Substitute \( t \) and \( dt \):
\( = \int \frac{dt}{4-t^2} = \int \frac{dt}{2^2-t^2} \)
Using the standard formula \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a}\log \left| \frac{a+x}{a-x} \right| + C \) with \( a=2 \):
\( = \frac{1}{2 \times 2}\log \left| \frac{2+t}{2-t} \right| + C \)
Substitute back \( t = \sin x \):
\( = \frac{1}{4}\log \left| \frac{2+\sin x}{2-\sin x} \right| + C \)
(iii) This integral is solved using integration by parts, treating \( \log(2+x^2) \) as \( u \) and \( 1 \) as \( dv \).
Let \( u = \log(2+x^2) \implies du = \frac{2x}{2+x^2} \, dx \). Let \( dv = 1 \, dx \implies v = x \).
\( I = x \log(2+x^2) - \int x \frac{2x}{2+x^2} \, dx \)
\( = x \log(2+x^2) - \int \frac{2x^2}{2+x^2} \, dx \)
For the remaining integral, we perform algebraic manipulation by adding and subtracting 2 in the numerator:
\( \int \frac{2x^2}{2+x^2} \, dx = \int \frac{2(x^2+2-2)}{x^2+2} \, dx = \int \left(2 - \frac{4}{x^2+2}\right) \, dx \)
Integrate term by term:
\( = \int 2 \, dx - 4 \int \frac{dx}{x^2+(\sqrt{2})^2} = 2x - 4 \times \frac{1}{\sqrt{2}}\tan^{-1} \frac{x}{\sqrt{2}} \)
Combine with the first part of integration by parts:
\( I = x \log(2+x^2) - \left(2x - \frac{4}{\sqrt{2}}\tan^{-1} \frac{x}{\sqrt{2}}\right) + C \)
\( = x \log(2+x^2) - 2x + 2\sqrt{2}\tan^{-1} \frac{x}{\sqrt{2}} + C \)
(iv) To solve this integral, we divide both the numerator and denominator by \( x^2 \).
\( \int \frac{x^2-4}{x^4+16} \, dx = \int \frac{1-\frac{4}{x^2}}{x^2+\frac{16}{x^2}} \, dx \)
Next, we use a substitution. Let \( u = x+\frac{4}{x} \). Then \( du = \left(1-\frac{4}{x^2}\right) \, dx \).
Squaring \( u \), we get \( u^2 = \left(x+\frac{4}{x}\right)^2 = x^2+\frac{16}{x^2}+8 \). So, \( x^2+\frac{16}{x^2} = u^2-8 \).
The integral transforms into:
\( = \int \frac{du}{u^2-8} = \int \frac{du}{u^2-(\sqrt{8})^2} \)
Using the formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a}\log \left| \frac{x-a}{x+a} \right| + C \) with \( a=\sqrt{8} = 2\sqrt{2} \):
\( = \frac{1}{2\sqrt{8}}\log \left| \frac{u-\sqrt{8}}{u+\sqrt{8}} \right| + C = \frac{1}{4\sqrt{2}}\log \left| \frac{u-2\sqrt{2}}{u+2\sqrt{2}} \right| + C \)
Substitute back \( u = x+\frac{4}{x} \):
\( = \frac{1}{4\sqrt{2}}\log \left| \frac{x+\frac{4}{x}-2\sqrt{2}}{x+\frac{4}{x}+2\sqrt{2}} \right| + C \)
Finally, simplify the terms inside the logarithm:
\( = \frac{1}{4\sqrt{2}}\log \left| \frac{x^2-2\sqrt{2}x+4}{x^2+2\sqrt{2}x+4} \right| + C \)
In simple words: These integrals required various advanced techniques, including splitting fractions, using integration by parts, and special substitutions. For integrals involving rational functions with \( x^4 \) or higher powers in the denominator, dividing by \( x^2 \) and then making a clever substitution like \( u = x \pm \frac{k}{x} \) is a common method to simplify them into standard forms.

🎯 Exam Tip: For rational functions where the degree of the numerator is less than the denominator, consider algebraic manipulation or partial fractions. When integrating \( \log x \), always use integration by parts with \( 1 \) as \( dv \). For specific forms like \( \int \frac{x^2 \pm k}{x^4+ax^2+b} \, dx \), dividing by \( x^2 \) is often the key initial step.

Question 8.
(i) \( \int \frac{x^2}{x^4+1} \, dx \)
(ii) \( \int \frac{dx}{x^4+1} \)
(iii) \( \int \frac{dx}{x^4+16} \)
(iv) \( \int \frac{x^2}{x^4+16} \, dx \)
Answer:
(i) To solve this integral, we first split it into two parts by adding and subtracting \( 1 \) in the numerator (after multiplying by 2).
\( I = \int \frac{x^2}{x^4+1} \, dx = \frac{1}{2}\int \frac{2x^2}{x^4+1} \, dx \)
\( = \frac{1}{2}\int \frac{(x^2+1)+(x^2-1)}{x^4+1} \, dx = \frac{1}{2}\int \frac{x^2+1}{x^4+1} \, dx + \frac{1}{2}\int \frac{x^2-1}{x^4+1} \, dx \)
Let \( I_1 = \int \frac{x^2+1}{x^4+1} \, dx \) and \( I_2 = \int \frac{x^2-1}{x^4+1} \, dx \). So, \( I = \frac{1}{2}I_1 + \frac{1}{2}I_2 \).
**For \( I_1 \):** We divide the numerator and denominator by \( x^2 \).
\( I_1 = \int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}} \, dx \)
Now, let \( t = x-\frac{1}{x} \). Then \( dt = \left(1+\frac{1}{x^2}\right) \, dx \).
Squaring \( t \), we get \( t^2 = x^2+\frac{1}{x^2}-2 \), so \( x^2+\frac{1}{x^2} = t^2+2 \).
Substituting these into \( I_1 \):
\( I_1 = \int \frac{dt}{t^2+2} = \int \frac{dt}{t^2+(\sqrt{2})^2} \)
Using the standard arctangent formula:
\( I_1 = \frac{1}{\sqrt{2}}\tan^{-1} \frac{t}{\sqrt{2}} + C_1 \)
Substitute back \( t = x-\frac{1}{x} \):
\( I_1 = \frac{1}{\sqrt{2}}\tan^{-1} \left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right) + C_1 = \frac{1}{\sqrt{2}}\tan^{-1} \left(\frac{x^2-1}{x\sqrt{2}}\right) + C_1 \)
**For \( I_2 \):** We also divide the numerator and denominator by \( x^2 \).
\( I_2 = \int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}} \, dx \)
This time, let \( u = x+\frac{1}{x} \). Then \( du = \left(1-\frac{1}{x^2}\right) \, dx \).
Squaring \( u \), we get \( u^2 = x^2+\frac{1}{x^2}+2 \), so \( x^2+\frac{1}{x^2} = u^2-2 \).
Substituting these into \( I_2 \):
\( I_2 = \int \frac{du}{u^2-2} = \int \frac{du}{u^2-(\sqrt{2})^2} \)
Using the standard logarithmic formula:
\( I_2 = \frac{1}{2\sqrt{2}}\log \left| \frac{u-\sqrt{2}}{u+\sqrt{2}} \right| + C_2 \)
Substitute back \( u = x+\frac{1}{x} \):
\( I_2 = \frac{1}{2\sqrt{2}}\log \left| \frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}} \right| + C_2 = \frac{1}{2\sqrt{2}}\log \left| \frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1} \right| + C_2 \)
(The final combination of \( I_1 \) and \( I_2 \) to get \( I \) is not provided within the specified page range.)
In simple words: These types of integrals are solved by a clever trick: splitting the fraction into two parts, then dividing both the top and bottom by \( x^2 \). This makes it possible to use specific substitutions like \( t = x - \frac{1}{x} \) or \( u = x + \frac{1}{x} \), which simplify the integral into standard forms that can be solved using arctangent or logarithmic formulas.

🎯 Exam Tip: For integrals of the form \( \int \frac{x^2 \pm 1}{x^4+k} \, dx \) or \( \int \frac{1}{x^4+k} \, dx \), dividing by \( x^2 \) and making substitutions \( t = x \mp \frac{1}{x} \) are essential techniques. Pay careful attention to the signs in the substitution, as they determine which standard formula to use (arctan vs. log).

Question 8.
(i) \( \int \frac{1}{x^2+36} d x \)
(ii) \( \int \frac{dx}{1+\frac{x^2}{4}} \)
(iii) \( \int \frac{dx}{50+2 x^2} \)
(iv) \( \int \frac{dx}{x^2-4} \)
(v) \( \int \frac{d x}{9 x^2-16} \)
(vi) \( \int \frac{d y}{25-16 y^2} \)
(vii) \( \int \frac{d y}{18-2 x^2} \)
Answer:
(i) Let \( I = \int \frac{1}{x^2+36} d x \). This can be rewritten as:
\( I = \int \frac{1}{x^2+6^2} d x \)
We use the formula \( \int \frac{d x}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \). Here, \( a=6 \).
\( I = \frac{1}{6}\tan^{-1} \frac{x}{6} + C \)
In simple words: This problem involves integrating a fraction with \( x^2 \) and a constant in the denominator. We use a standard integration formula that gives a tangent inverse function as the result. Just identify the value of 'a' and substitute it into the formula.

🎯 Exam Tip: Remember the basic integration formulas for inverse trigonometric functions. Being able to recognize the standard form \( \frac{1}{x^2+a^2} \) is crucial for quickly solving these types of problems.

Question 8.
(ii) \( \int \frac{dx}{1+\frac{x^2}{4}} \)
Answer: Let \( I = \int \frac{dx}{1+\frac{x^2}{4}} \).
First, simplify the denominator by finding a common denominator:
\( I = \int \frac{dx}{\frac{4+x^2}{4}} \)
\( I = \int \frac{4}{4+x^2} d x \)
\( I = \int \frac{4}{2^2+x^2} d x \)
We can take the constant 4 outside the integral:
\( I = 4 \int \frac{1}{x^2+2^2} d x \)
Using the formula \( \int \frac{d x}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \), where \( a=2 \):
\( I = 4 \left( \frac{1}{2} \tan^{-1} \frac{x}{2} \right) + C \)
\( I = 2 \tan^{-1} \frac{x}{2} + C \)
In simple words: Start by making the bottom part of the fraction simpler by removing the fraction inside it. Then, pull out any constant numbers and use the same formula as before. This helps in directly applying standard integration methods.

🎯 Exam Tip: Always simplify the integrand algebraically before applying any integration formulas. Look for common denominators or ways to factor out constants to match standard forms.

Question 8.
(iii) \( \int \frac{dx}{50+2 x^2} \)
Answer: Let \( I = \int \frac{d x}{50+2 x^2} \).
To match a standard form, we need the coefficient of \( x^2 \) to be 1. So, factor out 2 from the denominator:
\( I = \int \frac{d x}{2(25+x^2)} \)
Now, take \( \frac{1}{2} \) outside the integral:
\( I = \frac{1}{2}\int \frac{d x}{x^2+25} \)
This can be written as:
\( I = \frac{1}{2}\int \frac{d x}{x^2+5^2} \)
Using the formula \( \int \frac{d x}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \), where \( a=5 \):
\( I = \frac{1}{2} \times \frac{1}{5} \tan^{-1}\left(\frac{x}{5}\right) + C \)
\( I = \frac{1}{10} \tan^{-1} \left(\frac{x}{5}\right) + C \)
In simple words: When \( x^2 \) has a number in front, divide every part in the bottom by that number to make \( x^2 \) stand alone. Then, you can use the easy formula. This is a common first step in these types of integrals.

🎯 Exam Tip: Always ensure the coefficient of \( x^2 \) in the denominator is 1. Factor out any constant from the \( x^2 \) term to simplify the expression and match it with a standard integral form.

Question 8.
(iv) \( \int \frac{dx}{x^2-4} \)
Answer: Let \( I = \int \frac{dx}{x^2-4} \).
This can be written as:
\( I = \int \frac{dx}{x^2-2^2} \)
Using the formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \), where \( a=2 \):
\( I = \frac{1}{2 \times 2} \log \left| \frac{x-2}{x+2} \right| + C \)
\( I = \frac{1}{4} \log \left| \frac{x-2}{x+2} \right| + C \)
In simple words: For fractions like this, where the bottom is \( x^2 \) minus a number, use the logarithm formula. Just find the number that gets squared to give the constant, and then plug it into the formula. This formula is useful for integrals with a difference of squares in the denominator.

🎯 Exam Tip: Distinguish between \( x^2+a^2 \) (inverse tangent) and \( x^2-a^2 \) (logarithm). Pay close attention to the sign in the denominator to apply the correct formula.

Question 8.
(v) \( \int \frac{d x}{9 x^2-16} \)
Answer: Let \( I = \int \frac{d x}{9 x^2-16} \).
First, factor out 9 from the denominator to make the coefficient of \( x^2 \) equal to 1:
\( I = \int \frac{d x}{9\left(x^2-\frac{16}{9}\right)} \)
Take \( \frac{1}{9} \) outside the integral:
\( I = \frac{1}{9}\int \frac{d x}{x^2-\left(\frac{4}{3}\right)^2} \)
Using the formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \), where \( a=\frac{4}{3} \):
\( I = \frac{1}{9} \times \frac{1}{2 \times \frac{4}{3}} \log \left| \frac{x-\frac{4}{3}}{x+\frac{4}{3}} \right| + C \)
Simplify the constant part:
\( I = \frac{1}{9} \times \frac{3}{8} \log \left| \frac{\frac{3x-4}{3}}{\frac{3x+4}{3}} \right| + C \)
\( I = \frac{1}{24} \log \left| \frac{3x-4}{3x+4} \right| + C \)
In simple words: Like before, make sure \( x^2 \) is by itself on the bottom. Then, change the constant number into something squared. After that, use the logarithm formula. Remember to simplify fractions in the logarithm.

🎯 Exam Tip: Be careful with fractions when identifying 'a' and simplifying the constant \( \frac{1}{2a} \) term. A common mistake is not simplifying the fraction inside the logarithm correctly.

Question 8.
(vi) \( \int \frac{d y}{25-16 y^2} \)
Answer: Let \( I = \int \frac{d y}{25-16 y^2} \).
Factor out 16 from the denominator to get \( y^2 \) with a coefficient of 1:
\( I = \int \frac{d y}{16\left(\frac{25}{16}-y^2\right)} \)
Take \( \frac{1}{16} \) outside the integral:
\( I = \frac{1}{16}\int \frac{d y}{\left(\frac{5}{4}\right)^2-y^2} \)
Using the formula \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \), where \( a=\frac{5}{4} \) and the variable is \( y \):
\( I = \frac{1}{16} \times \frac{1}{2 \times \frac{5}{4}} \log \left| \frac{\frac{5}{4}+y}{\frac{5}{4}-y} \right| + C \)
Simplify the constant term:
\( I = \frac{1}{16} \times \frac{4}{10} \log \left| \frac{5+4y}{5-4y} \right| + C \)
\( I = \frac{1}{40} \log \left| \frac{5+4y}{5-4y} \right| + C \)
In simple words: This integral is similar to others, but the variable is \( y \) and the constant term is first. Pull out the number with \( y^2 \), rewrite the constant as a square, and then use the appropriate logarithm formula, being careful with the signs.

🎯 Exam Tip: Pay close attention to the order of terms in the denominator. If it's \( a^2-x^2 \), the formula is \( \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| \). If it's \( x^2-a^2 \), the formula is \( \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| \). A common error is mixing these two up.

Question 8.
(vii) \( \int \frac{d y}{18-2 x^2} \)
Answer: Let \( I = \int \frac{d y}{18-2 x^2} \).
First, factor out 2 from the denominator:
\( I = \int \frac{d x}{2(9-x^2)} \)
Take \( \frac{1}{2} \) outside the integral:
\( I = \frac{1}{2}\int \frac{d x}{3^2-x^2} \)
Using the formula \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \), where \( a=3 \):
\( I = \frac{1}{2} \times \frac{1}{2 \times 3} \log \left| \frac{3+x}{3-x} \right| + C \)
\( I = \frac{1}{12} \log \left| \frac{3+x}{3-x} \right| + C \)
In simple words: Again, take out the number from the denominator so \( x^2 \) is left alone. Then, use the formula for integrals that have a constant squared minus \( x^2 \). This method is very systematic.

🎯 Exam Tip: Always look to simplify the denominator by factoring out common constants, which makes it easier to identify the correct standard integral formula to use.

Question 9.
(i) \( \int \frac{x^2+1}{x^4+x^2+1} d x \)
Answer: Let \( I = \int \frac{x^2+1}{x^4+x^2+1} d x \).
To simplify this integral, divide both the numerator and the denominator by \( x^2 \):
\( I = \int \frac{\frac{x^2}{x^2}+\frac{1}{x^2}}{\frac{x^4}{x^2}+\frac{x^2}{x^2}+\frac{1}{x^2}} dx \)
\( I = \int \frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}} dx \)
We can rewrite the denominator as \( \left(x-\frac{1}{x}\right)^2+3 \) or \( \left(x+\frac{1}{x}\right)^2-1 \). Let's use the substitution related to \( \left(x-\frac{1}{x}\right) \).
Let \( x - \frac{1}{x} = t \).
Then, differentiate both sides with respect to \( x \):
\( \left(1 - \left(-\frac{1}{x^2}\right)\right) dx = dt \)
\( \left(1 + \frac{1}{x^2}\right) dx = dt \)
Also, from \( x - \frac{1}{x} = t \), square both sides:
\( \left(x-\frac{1}{x}\right)^2 = t^2 \)
\( x^2 + \frac{1}{x^2} - 2 = t^2 \)
\( x^2 + \frac{1}{x^2} = t^2 + 2 \)
Now substitute these into the integral:
\( I = \int \frac{dt}{t^2+2+1} \)
\( I = \int \frac{dt}{t^2+(\sqrt{3})^2} \)
Using the formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \), where \( a=\sqrt{3} \):
\( I = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C_1 \)
Substitute back \( t = x - \frac{1}{x} \):
\( I = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x - \frac{1}{x}}{\sqrt{3}}\right) + C_1 \)
\( I = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) + C_1 \)
In simple words: For fractions with \( x^4 \) in the bottom, a good trick is to divide everything by \( x^2 \). Then, look for a substitution like \( x - \frac{1}{x} = t \) or \( x + \frac{1}{x} = t \). This turns the problem into a simpler integral form that uses \( \tan^{-1} \).

🎯 Exam Tip: When dealing with integrals of the form \( \frac{x^2 \pm 1}{x^4+kx^2+1} \), dividing numerator and denominator by \( x^2 \) is a standard technique. This transforms the integral into a form solvable by substitution and standard inverse trigonometric formulas. Remember to handle both the numerator and the denominator correctly during substitution.

Question 9.
(ii) \( \int \frac{x^2+9}{x^4-2 x^2+81} d x \)
Answer: Let \( I = \int \frac{x^2+9}{x^4-2 x^2+81} d x \).
Divide both the numerator and the denominator by \( x^2 \):
\( I = \int \frac{\frac{x^2}{x^2}+\frac{9}{x^2}}{\frac{x^4}{x^2}-\frac{2x^2}{x^2}+\frac{81}{x^2}} dx \)
\( I = \int \frac{1+\frac{9}{x^2}}{x^2-2+\frac{81}{x^2}} dx \)
Now, use the substitution \( x - \frac{9}{x} = t \).
Differentiate both sides with respect to \( x \):
\( \left(1 - \left(-\frac{9}{x^2}\right)\right) dx = dt \)
\( \left(1 + \frac{9}{x^2}\right) dx = dt \)
Also, from \( x - \frac{9}{x} = t \), square both sides:
\( \left(x-\frac{9}{x}\right)^2 = t^2 \)
\( x^2 + \frac{81}{x^2} - 2(x)\left(\frac{9}{x}\right) = t^2 \)
\( x^2 + \frac{81}{x^2} - 18 = t^2 \)
\( x^2 + \frac{81}{x^2} = t^2 + 18 \)
Substitute these into the integral:
\( I = \int \frac{dt}{t^2+18-2} \)
\( I = \int \frac{dt}{t^2+16} \)
\( I = \int \frac{dt}{t^2+4^2} \)
Using the formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \), where \( a=4 \):
\( I = \frac{1}{4} \tan^{-1}\left(\frac{t}{4}\right) + C \)
Substitute back \( t = x - \frac{9}{x} \):
\( I = \frac{1}{4} \tan^{-1}\left(\frac{x - \frac{9}{x}}{4}\right) + C \)
\( I = \frac{1}{4} \tan^{-1}\left(\frac{x^2-9}{4x}\right) + C \)
In simple words: Follow the same steps as the previous problem: divide by \( x^2 \), pick the right substitution, and then solve the simpler integral. The key is setting up the substitution so the numerator becomes \( dt \) and the denominator simplifies.

🎯 Exam Tip: For expressions like \( x^2-2+\frac{81}{x^2} \), identify if it's part of \( (x - \frac{k}{x})^2 \) or \( (x + \frac{k}{x})^2 \). In this case, since the numerator has \( 1+\frac{9}{x^2} \), the substitution \( x - \frac{9}{x} = t \) is appropriate as its derivative matches the numerator.

Question 9.
(iii) \( \int \frac{x^2-8}{x^4+7 x^2+64} d x \)
Answer: Let \( I = \int \frac{x^2-8}{x^4+7 x^2+64} d x \).
Divide both the numerator and the denominator by \( x^2 \):
\( I = \int \frac{\frac{x^2}{x^2}-\frac{8}{x^2}}{\frac{x^4}{x^2}+\frac{7x^2}{x^2}+\frac{64}{x^2}} dx \)
\( I = \int \frac{1-\frac{8}{x^2}}{x^2+7+\frac{64}{x^2}} dx \)
Now, use the substitution \( x + \frac{8}{x} = t \).
Differentiate both sides with respect to \( x \):
\( \left(1 - \frac{8}{x^2}\right) dx = dt \)
Also, from \( x + \frac{8}{x} = t \), square both sides:
\( \left(x+\frac{8}{x}\right)^2 = t^2 \)
\( x^2 + \frac{64}{x^2} + 2(x)\left(\frac{8}{x}\right) = t^2 \)
\( x^2 + \frac{64}{x^2} + 16 = t^2 \)
\( x^2 + \frac{64}{x^2} = t^2 - 16 \)
Substitute these into the integral:
\( I = \int \frac{dt}{t^2-16+7} \)
\( I = \int \frac{dt}{t^2-9} \)
\( I = \int \frac{dt}{t^2-3^2} \)
Using the formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \), where \( a=3 \):
\( I = \frac{1}{2 \times 3} \log \left| \frac{t-3}{t+3} \right| + C \)
\( I = \frac{1}{6} \log \left| \frac{x+\frac{8}{x}-3}{x+\frac{8}{x}+3} \right| + C \)
\( I = \frac{1}{6} \log \left| \frac{x^2-3x+8}{x^2+3x+8} \right| + C \)
In simple words: This problem again uses the division by \( x^2 \) trick. Since the numerator has \( 1-\frac{8}{x^2} \), we choose the substitution \( x+\frac{8}{x}=t \). This type of substitution simplifies the integral to a basic logarithmic form.

🎯 Exam Tip: The choice of substitution (either \( x - \frac{k}{x} \) or \( x + \frac{k}{x} \)) depends on the numerator. If the numerator is \( 1+\frac{k}{x^2} \), use \( x - \frac{k}{x} = t \). If it's \( 1-\frac{k}{x^2} \), use \( x + \frac{k}{x} = t \). This ensures that the derivative matches the simplified numerator.

Question 10.
(i) \( \int \frac{x^2}{x^4+x^2+1} d x \)
Answer: Let \( I = \int \frac{x^2}{x^4+x^2+1} d x \).
To solve this, multiply and divide by 2:
\( I = \frac{1}{2}\int \frac{2 x^2}{x^4+x^2+1} d x \)
Now, rewrite the numerator \( 2x^2 \) as \( (x^2+1) + (x^2-1) \):
\( I = \frac{1}{2}\int \frac{(x^2+1)+(x^2-1)}{x^4+x^2+1} d x \)
Split this into two separate integrals:
\( I = \frac{1}{2}\int \frac{x^2+1}{x^4+x^2+1} d x + \frac{1}{2}\int \frac{x^2-1}{x^4+x^2+1} d x \)
Let \( I_1 = \int \frac{x^2+1}{x^4+x^2+1} d x \) and \( I_2 = \int \frac{x^2-1}{x^4+x^2+1} d x \). So \( I = \frac{1}{2}I_1 + \frac{1}{2}I_2 \).
For \( I_1 \), divide the numerator and denominator by \( x^2 \):
\( I_1 = \int \frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}} dx \)
Let \( x - \frac{1}{x} = t \).
Then \( \left(1 + \frac{1}{x^2}\right)dx = dt \).
Also, \( \left(x-\frac{1}{x}\right)^2 = t^2 \)
\( x^2 + \frac{1}{x^2} - 2 = t^2 \)
\( x^2 + \frac{1}{x^2} = t^2 + 2 \)
Substitute these into \( I_1 \):
\( I_1 = \int \frac{dt}{t^2+2+1} \)
\( I_1 = \int \frac{dt}{t^2+(\sqrt{3})^2} \)
Using the formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \), where \( a=\sqrt{3} \):
\( I_1 = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C_1 \)
Substitute back \( t = x - \frac{1}{x} \):
\( I_1 = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) + C_1 \)
For \( I_2 \), divide the numerator and denominator by \( x^2 \):
\( I_2 = \int \frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}} dx \)
Let \( x + \frac{1}{x} = u \).
Then \( \left(1 - \frac{1}{x^2}\right)dx = du \).
Also, \( \left(x+\frac{1}{x}\right)^2 = u^2 \)
\( x^2 + \frac{1}{x^2} + 2 = u^2 \)
\( x^2 + \frac{1}{x^2} = u^2 - 2 \)
Substitute these into \( I_2 \):
\( I_2 = \int \frac{du}{u^2-2+1} \)
\( I_2 = \int \frac{du}{u^2-1^2} \)
Using the formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \), where \( a=1 \):
\( I_2 = \frac{1}{2 \times 1} \log \left| \frac{u-1}{u+1} \right| + C_2 \)
\( I_2 = \frac{1}{2} \log \left| \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} \right| + C_2 \)
\( I_2 = \frac{1}{2} \log \left| \frac{x^2-x+1}{x^2+x+1} \right| + C_2 \)
Now, combine \( I_1 \) and \( I_2 \) to find \( I \):
\( I = \frac{1}{2} \left( \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) + C_1 \right) + \frac{1}{2} \left( \frac{1}{2} \log \left| \frac{x^2-x+1}{x^2+x+1} \right| + C_2 \right) \)
\( I = \frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) + \frac{1}{4} \log \left| \frac{x^2-x+1}{x^2+x+1} \right| + C \), where \( C = \frac{C_1}{2} + \frac{C_2}{2} \).
In simple words: This integral is solved by first splitting it into two parts: one with \( x^2+1 \) in the numerator and one with \( x^2-1 \). Each part is then solved using a different substitution (either \( x-\frac{1}{x} \) or \( x+\frac{1}{x} \)) and a standard formula. Finally, the results from both parts are added together.

🎯 Exam Tip: Integrals of the form \( \frac{x^2}{x^4+kx^2+1} \) often require splitting the numerator into \( (x^2+1) \) and \( (x^2-1) \) parts after multiplying by a constant. This strategy allows the use of two different substitutions that lead to inverse tangent and logarithmic forms, respectively.

Question 10.
(ii) \( \int \frac{dx}{x^4+x^2+1} \)
Answer: Let \( I = \int \frac{d x}{x^4+x^2+1} \).
To solve this, multiply and divide by 2:
\( I = \frac{1}{2}\int \frac{2 d x}{x^4+x^2+1} d x \)
Now, rewrite the numerator \( 2 \) as \( (x^2+1) - (x^2-1) \):
\( I = \frac{1}{2}\int \frac{(x^2+1)-(x^2-1)}{x^4+x^2+1} d x \)
Split this into two separate integrals:
\( I = \frac{1}{2}\int \frac{x^2+1}{x^4+x^2+1} d x - \frac{1}{2}\int \frac{x^2-1}{x^4+x^2+1} d x \)
Let \( I_1 = \int \frac{x^2+1}{x^4+x^2+1} d x \) and \( I_2 = \int \frac{x^2-1}{x^4+x^2+1} d x \). So \( I = \frac{1}{2}I_1 - \frac{1}{2}I_2 \).
For \( I_1 \), divide the numerator and denominator by \( x^2 \):
\( I_1 = \int \frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}} dx \)
Let \( x - \frac{1}{x} = t \).
Then \( \left(1 + \frac{1}{x^2}\right)dx = dt \).
Also, \( \left(x-\frac{1}{x}\right)^2 = t^2 \)
\( x^2 + \frac{1}{x^2} - 2 = t^2 \)
\( x^2 + \frac{1}{x^2} = t^2 + 2 \)
Substitute these into \( I_1 \):
\( I_1 = \int \frac{dt}{t^2+2+1} \)
\( I_1 = \int \frac{dt}{t^2+(\sqrt{3})^2} \)
Using the formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \), where \( a=\sqrt{3} \):
\( I_1 = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C_1 \)
Substitute back \( t = x - \frac{1}{x} \):
\( I_1 = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) + C_1 \)
For \( I_2 \), divide the numerator and denominator by \( x^2 \):
\( I_2 = \int \frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}} dx \)
Let \( x + \frac{1}{x} = u \).
Then \( \left(1 - \frac{1}{x^2}\right)dx = du \).
Also, \( \left(x+\frac{1}{x}\right)^2 = u^2 \)
\( x^2 + \frac{1}{x^2} + 2 = u^2 \)
\( x^2 + \frac{1}{x^2} = u^2 - 2 \)
Substitute these into \( I_2 \):
\( I_2 = \int \frac{du}{u^2-2+1} \)
\( I_2 = \int \frac{du}{u^2-1^2} \)
Using the formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \), where \( a=1 \):
\( I_2 = \frac{1}{2 \times 1} \log \left| \frac{u-1}{u+1} \right| + C_2 \)
\( I_2 = \frac{1}{2} \log \left| \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} \right| + C_2 \)
\( I_2 = \frac{1}{2} \log \left| \frac{x^2-x+1}{x^2+x+1} \right| + C_2 \)
Now, combine \( I_1 \) and \( I_2 \) to find \( I \):
\( I = \frac{1}{2} \left( \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) + C_1 \right) - \frac{1}{2} \left( \frac{1}{2} \log \left| \frac{x^2-x+1}{x^2+x+1} \right| + C_2 \right) \)
\( I = \frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) - \frac{1}{4} \log \left| \frac{x^2-x+1}{x^2+x+1} \right| + C \), where \( C = \frac{C_1}{2} - \frac{C_2}{2} \).
In simple words: This integral is very similar to the previous one, but the numerator of the original integral is a constant. We add and subtract \( x^2 \) in the numerator to create terms of \( x^2+1 \) and \( x^2-1 \). This allows us to split the integral into two parts, each solved with a specific substitution, leading to an inverse tangent and a logarithmic result.

🎯 Exam Tip: For integrals of the form \( \frac{1}{x^4+kx^2+1} \), the technique is to manipulate the numerator as \( \frac{1}{2}((x^2+1) - (x^2-1)) \) after multiplying by 2. This creates two distinct integrals that can be solved using the standard substitutions \( x-\frac{1}{x} \) and \( x+\frac{1}{x} \).

Question 11. \( \int \frac{(x-1)^2}{x^4+x^2+1} d x \)
Answer:
Let \( I = \int \frac{(x-1)^2}{x^4+x^2+1} d x \)
First, expand the numerator: \( (x-1)^2 = x^2 - 2x + 1 \)
So, \( I = \int \frac{x^2 - 2x + 1}{x^4+x^2+1} d x \)
We can split this into two parts by adding and subtracting terms in the numerator to match the denominator structure.
This integral can be broken down into two simpler integrals, \( I_1 \) and \( I_2 \), for easier calculation.
\( I = \int \frac{x^2+1}{x^4+x^2+1} d x - \int \frac{2x}{x^4+x^2+1} d x \)
\( I = I_1 - I_2 \)

For \( I_1 = \int \frac{x^2+1}{x^4+x^2+1} d x \):
Divide the numerator and denominator by \( x^2 \):
\( I_1 = \int \frac{1 + \frac{1}{x^2}}{x^2+1+\frac{1}{x^2}} d x \)
Rearrange the denominator: \( x^2+1+\frac{1}{x^2} = (x-\frac{1}{x})^2 + 2 + 1 = (x-\frac{1}{x})^2 + (\sqrt{3})^2 \)
Let \( t = x - \frac{1}{x} \).
Then \( dt = (1 + \frac{1}{x^2}) d x \).
So, \( I_1 = \int \frac{dt}{t^2+(\sqrt{3})^2} \)
We use the standard integral formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C \). Here, \( a = \sqrt{3} \).
\( I_1 = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C_1 \)
Substitute \( t = x - \frac{1}{x} = \frac{x^2-1}{x} \) back:
\( I_1 = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) + C_1 \)

For \( I_2 = \int \frac{2x}{x^4+x^2+1} d x \):
Let \( u = x^2 \).
Then \( du = 2x d x \).
So, \( I_2 = \int \frac{du}{u^2+u+1} \)
Complete the square in the denominator: \( u^2+u+1 = (u^2+u+\frac{1}{4}) + 1 - \frac{1}{4} = (u+\frac{1}{2})^2 + \frac{3}{4} = (u+\frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \)
So, \( I_2 = \int \frac{du}{(u+\frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \)
Again, using the formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C \). Here, \( a = \frac{\sqrt{3}}{2} \).
\( I_2 = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{u+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C_2 \)
\( I_2 = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{\frac{2u+1}{2}}{\frac{\sqrt{3}}{2}}\right) + C_2 \)
\( I_2 = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2u+1}{\sqrt{3}}\right) + C_2 \)
Substitute \( u = x^2 \) back:
\( I_2 = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right) + C_2 \)

Finally, combine \( I_1 \) and \( I_2 \):
\( I = I_1 - I_2 \)
\( I = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right) + C \)
Here, \( C = C_1 - C_2 \) is the constant of integration. This integral requires careful splitting and substitution to solve effectively.
In simple words: First, we break the original fraction into two parts that are easier to integrate. For each part, we use a trick called substitution and a common integral formula. We put \( x - \frac{1}{x} \) as 't' for the first part and \( x^2 \) as 'u' for the second part, then solve each one separately. Finally, we combine the answers from both parts to get the total solution.

🎯 Exam Tip: When integrating complex rational functions, look for ways to split the integral into sums or differences of simpler forms. Completing the square in the denominator and appropriate substitutions (like \( t = x - \frac{1}{x} \) or \( t = x + \frac{1}{x} \)) are key techniques for integrals involving \( x^4+ax^2+b \).