RBSE Solutions Class 12 Maths Chapter 9 Integration Exercise 9.5

Exercise 9.5: Integrate the following functions with respect to x

Question 1. \( \frac{1}{x^2+2x+10} \)
Answer: To integrate this function, we first complete the square in the denominator. We add and subtract 1 (which is \( (2/2)^2 \)) to \( x^2+2x \). This helps us write the denominator in the form \( (x+a)^2 + b^2 \).
\[ \int \frac{1}{x^2+2x+10} \, dx \] \[ = \int \frac{1}{x^2+2x \cdot 1 \cdot x + 1^2 + 9} \, dx \] \[ = \int \frac{1}{(x+1)^2 + 3^2} \, dx \] Using the standard integral formula \( \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), where here \( x \) is \( (x+1) \) and \( a \) is 3.
\[ = \frac{1}{3} \tan^{-1} \left( \frac{x+1}{3} \right) + C \]In simple words: We changed the bottom part of the fraction to a squared term plus a constant. Then we used a special formula for integrals to find the answer.

🎯 Exam Tip: Remember to complete the square in the denominator for quadratic expressions before applying standard integral formulas like \( \int \frac{1}{x^2+a^2} \, dx \) or \( \int \frac{1}{x^2-a^2} \, dx \).

Question 3. \( \frac{1}{9x^2 - 12x + 8} \)
Answer: We begin by taking out the common factor of 9 from the denominator to simplify the quadratic expression. Then, we complete the square for the remaining quadratic term. This allows us to use a standard integration formula.
\[ \int \frac{1}{9x^2-12x+8} \, dx \] \[ = \int \frac{1}{9 \left( x^2 - \frac{12}{9}x + \frac{8}{9} \right)} \, dx \] \[ = \frac{1}{9} \int \frac{1}{x^2 - \frac{4}{3}x + \frac{8}{9}} \, dx \] To complete the square for \( x^2 - \frac{4}{3}x \), we add and subtract \( \left( \frac{4/3}{2} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \).
\[ = \frac{1}{9} \int \frac{1}{x^2 - 2 \cdot x \cdot \frac{2}{3} + \left( \frac{2}{3} \right)^2 - \frac{4}{9} + \frac{8}{9}} \, dx \] \[ = \frac{1}{9} \int \frac{1}{\left( x - \frac{2}{3} \right)^2 + \frac{4}{9}} \, dx \] \[ = \frac{1}{9} \int \frac{1}{\left( x - \frac{2}{3} \right)^2 + \left( \frac{2}{3} \right)^2} \, dx \] Using the formula \( \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), with \( x = \left( x - \frac{2}{3} \right) \) and \( a = \frac{2}{3} \).
\[ = \frac{1}{9} \cdot \frac{1}{\frac{2}{3}} \tan^{-1} \left( \frac{x - \frac{2}{3}}{\frac{2}{3}} \right) + C \] \[ = \frac{1}{9} \cdot \frac{3}{2} \tan^{-1} \left( \frac{\frac{3x-2}{3}}{\frac{2}{3}} \right) + C \] \[ = \frac{1}{6} \tan^{-1} \left( \frac{3x-2}{2} \right) + C \]In simple words: First, we took out 9 from the bottom part. Then, we made the bottom part into a perfect square plus a number. Finally, we used the tangent inverse formula to solve the integral.

🎯 Exam Tip: Always factor out the coefficient of \( x^2 \) if it's not 1 before completing the square to avoid errors in the `a` term of the integral formula.

Question 4. \( \frac{1}{3+2x-x^2} \)
Answer: For this integral, we again use the method of completing the square in the denominator. Since the \( x^2 \) term is negative, we factor out -1, then rearrange and complete the square to get a form like \( a^2 - (x \pm b)^2 \).
\[ \int \frac{1}{3+2x-x^2} \, dx \] \[ = \int \frac{1}{3 - (x^2-2x)} \, dx \] To complete the square for \( x^2-2x \), we add and subtract \( (2/2)^2 = 1^2 = 1 \).
\[ = \int \frac{1}{3 - (x^2-2x+1-1)} \, dx \] \[ = \int \frac{1}{3 - ((x-1)^2 - 1)} \, dx \] \[ = \int \frac{1}{3 - (x-1)^2 + 1} \, dx \] \[ = \int \frac{1}{4 - (x-1)^2} \, dx \] \[ = \int \frac{1}{2^2 - (x-1)^2} \, dx \] Using the formula \( \int \frac{1}{a^2-x^2} \, dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \), with \( a=2 \) and \( x=(x-1) \).
\[ = \frac{1}{2 \cdot 2} \log \left| \frac{2+(x-1)}{2-(x-1)} \right| + C \] \[ = \frac{1}{4} \log \left| \frac{2+x-1}{2-x+1} \right| + C \] \[ = \frac{1}{4} \log \left| \frac{x+1}{3-x} \right| + C \]In simple words: We rewrite the bottom part by making a perfect square. Because of the minus sign in front of \( x^2 \), we change the form to a constant minus a squared term. Then we use a specific logarithm formula for integrals.

🎯 Exam Tip: When \( x^2 \) has a negative coefficient, be very careful with the signs when completing the square and applying the correct integral formula (e.g., \( a^2-x^2 \) vs. \( x^2-a^2 \)).

Question 6. \( \frac{\cos x}{\sin^2 x + 4 \sin x + 5} \)
Answer: To solve this integral, we first use a substitution. Let \( \sin x = t \), then the derivative \( \cos x \, dx \) becomes \( dt \). This simplifies the integral into a standard form.
Let \( t = \sin x \). Then \( dt = \cos x \, dx \). The integral becomes:
\[ \int \frac{dt}{t^2 + 4t + 5} \] Now, we complete the square in the denominator: \( t^2+4t+5 = t^2+4t+4+1 = (t+2)^2+1^2 \).
\[ = \int \frac{dt}{(t+2)^2 + 1^2} \] Using the standard integral formula \( \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), with \( x = (t+2) \) and \( a=1 \).
\[ = \frac{1}{1} \tan^{-1} \left( \frac{t+2}{1} \right) + C \] \[ = \tan^{-1} (t+2) + C \] Finally, substitute back \( t = \sin x \).
\[ = \tan^{-1} (\sin x + 2) + C \]In simple words: We change \( \sin x \) to \( t \) to make the integral simpler. After that, we make the bottom part a perfect square and use the tangent inverse formula. Then, we put \( \sin x \) back into the answer.

🎯 Exam Tip: For integrals involving trigonometric functions, look for suitable substitutions that transform the integral into a standard algebraic form, often involving completing the square.

Question 7. \( \frac{x-3}{x^2+2x-4} \)
Answer: This integral is of the form \( \int \frac{px+q}{ax^2+bx+c} \, dx \). We express the numerator as a linear combination of the derivative of the denominator and a constant.
Let the numerator \( x-3 = A \frac{d}{dx}(x^2+2x-4) + B \).
\( x-3 = A(2x+2) + B \)
\( x-3 = 2Ax + 2A + B \) Comparing coefficients of \( x \):
\( 2A = 1 \implies A = \frac{1}{2} \) Comparing constant terms:
\( 2A+B = -3 \)
\( 2\left(\frac{1}{2}\right) + B = -3 \)
\( 1+B = -3 \implies B = -4 \) So, \( x-3 = \frac{1}{2}(2x+2) - 4 \). Now, substitute this back into the integral:
\[ \int \frac{x-3}{x^2+2x-4} \, dx = \int \frac{\frac{1}{2}(2x+2) - 4}{x^2+2x-4} \, dx \] We split this into two simpler integrals.
\[ = \frac{1}{2} \int \frac{2x+2}{x^2+2x-4} \, dx - 4 \int \frac{1}{x^2+2x-4} \, dx \] For the first integral, let \( u = x^2+2x-4 \), then \( du = (2x+2) \, dx \). So, \( \int \frac{du}{u} = \log|u| \).
\[ = \frac{1}{2} \log|x^2+2x-4| - 4 \int \frac{1}{x^2+2x-4} \, dx \] For the second integral, we complete the square in the denominator: \( x^2+2x-4 = x^2+2x+1-1-4 = (x+1)^2-5 = (x+1)^2-(\sqrt{5})^2 \).
\[ = \frac{1}{2} \log|x^2+2x-4| - 4 \int \frac{1}{(x+1)^2-(\sqrt{5})^2} \, dx \] Using the formula \( \int \frac{1}{x^2-a^2} \, dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C \), with \( x=(x+1) \) and \( a=\sqrt{5} \).
\[ = \frac{1}{2} \log|x^2+2x-4| - 4 \left( \frac{1}{2\sqrt{5}} \log \left| \frac{x+1-\sqrt{5}}{x+1+\sqrt{5}} \right| \right) + C \]
\[ = \frac{1}{2} \log|x^2+2x-4| - \frac{2}{\sqrt{5}} \log \left| \frac{x+1-\sqrt{5}}{x+1+\sqrt{5}} \right| + C \]In simple words: First, we write the top part of the fraction in terms of the derivative of the bottom part. This helps us split the problem into two easier integrals. Then, we solve each part separately, one with a logarithm and the other by completing the square and using another logarithm formula.

🎯 Exam Tip: When integrating fractions where the numerator is a linear function and the denominator is a quadratic, always try to express the numerator using the derivative of the denominator to simplify the integral into known forms.

Question 8. \( \frac{3x+1}{2x^2-2x+3} \)
Answer: We use a similar strategy as in the previous question, expressing the numerator in terms of the derivative of the denominator.
Let \( 3x+1 = A \frac{d}{dx}(2x^2-2x+3) + B \).
\( 3x+1 = A(4x-2) + B \)
\( 3x+1 = 4Ax - 2A + B \) Comparing coefficients of \( x \):
\( 4A = 3 \implies A = \frac{3}{4} \) Comparing constant terms:
\( -2A+B = 1 \)
\( -2\left(\frac{3}{4}\right) + B = 1 \)
\( -\frac{3}{2} + B = 1 \implies B = 1 + \frac{3}{2} = \frac{5}{2} \) So, \( 3x+1 = \frac{3}{4}(4x-2) + \frac{5}{2} \). Now, substitute this back into the integral:
\[ \int \frac{3x+1}{2x^2-2x+3} \, dx = \int \frac{\frac{3}{4}(4x-2) + \frac{5}{2}}{2x^2-2x+3} \, dx \] Split this into two integrals.
\[ = \frac{3}{4} \int \frac{4x-2}{2x^2-2x+3} \, dx + \frac{5}{2} \int \frac{1}{2x^2-2x+3} \, dx \] For the first integral, let \( u = 2x^2-2x+3 \), then \( du = (4x-2) \, dx \). So, \( \int \frac{du}{u} = \log|u| \).
\[ = \frac{3}{4} \log|2x^2-2x+3| + \frac{5}{2} \int \frac{1}{2x^2-2x+3} \, dx \] For the second integral, factor out 2 from the denominator and complete the square:
\( 2x^2-2x+3 = 2(x^2-x+\frac{3}{2}) \) To complete the square for \( x^2-x \), we add and subtract \( (-1/2)^2 = 1/4 \).
\( 2\left(x^2-x+\frac{1}{4}-\frac{1}{4}+\frac{3}{2}\right) = 2\left(\left(x-\frac{1}{2}\right)^2 + \frac{5}{4}\right) = 2\left(\left(x-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2\right) \) So the second integral becomes:
\[ \frac{5}{2} \int \frac{1}{2\left(\left(x-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2\right)} \, dx \] \[ = \frac{5}{4} \int \frac{1}{\left(x-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{5}}{2}\right)^2} \, dx \] Using the formula \( \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), with \( x=\left(x-\frac{1}{2}\right) \) and \( a=\frac{\sqrt{5}}{2} \).
\[ = \frac{5}{4} \left( \frac{1}{\frac{\sqrt{5}}{2}} \tan^{-1} \left( \frac{x-\frac{1}{2}}{\frac{\sqrt{5}}{2}} \right) \right) + C_2 \] \[ = \frac{5}{4} \cdot \frac{2}{\sqrt{5}} \tan^{-1} \left( \frac{2x-1}{\sqrt{5}} \right) + C_2 \] \[ = \frac{\sqrt{5}}{2} \tan^{-1} \left( \frac{2x-1}{\sqrt{5}} \right) + C_2 \] Combining both parts:
\[ \int \frac{3x+1}{2x^2-2x+3} \, dx = \frac{3}{4} \log|2x^2-2x+3| + \frac{\sqrt{5}}{2} \tan^{-1} \left( \frac{2x-1}{\sqrt{5}} \right) + C \] Where \( C = C_1 + C_2 \).In simple words: We rewrite the top part using the derivative of the bottom part, which splits the integral into two parts. The first part is a simple logarithm. For the second part, we complete the square in the denominator and use the inverse tangent formula.

🎯 Exam Tip: Remember to carry through the coefficient of the \( x^2 \) term when completing the square, as it affects the 'a' value in the standard integral formulas and may need to be factored out. This often requires careful algebraic manipulation.

Question 9. \( \frac{x+1}{x^2+4x+5} \)
Answer: We use the same method as before for integrals of this type. We express the numerator as a combination of the derivative of the denominator and a constant.
Let \( x+1 = A \frac{d}{dx}(x^2+4x+5) + B \).
\( x+1 = A(2x+4) + B \)
\( x+1 = 2Ax + 4A + B \) Comparing coefficients of \( x \):
\( 2A = 1 \implies A = \frac{1}{2} \) Comparing constant terms:
\( 4A+B = 1 \)
\( 4\left(\frac{1}{2}\right) + B = 1 \)
\( 2+B = 1 \implies B = -1 \) So, \( x+1 = \frac{1}{2}(2x+4) - 1 \). Substitute this into the integral:
\[ \int \frac{x+1}{x^2+4x+5} \, dx = \int \frac{\frac{1}{2}(2x+4) - 1}{x^2+4x+5} \, dx \] Split this into two integrals:
\[ = \frac{1}{2} \int \frac{2x+4}{x^2+4x+5} \, dx - \int \frac{1}{x^2+4x+5} \, dx \] For the first integral, let \( u = x^2+4x+5 \), so \( du = (2x+4) \, dx \). Thus, it integrates to \( \log|u| \).
\[ = \frac{1}{2} \log|x^2+4x+5| - \int \frac{1}{x^2+4x+5} \, dx \] For the second integral, complete the square in the denominator: \( x^2+4x+5 = x^2+4x+4+1 = (x+2)^2+1^2 \).
\[ = \frac{1}{2} \log|x^2+4x+5| - \int \frac{1}{(x+2)^2+1^2} \, dx \] Using the formula \( \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), with \( x=(x+2) \) and \( a=1 \).
\[ = \frac{1}{2} \log|x^2+4x+5| - \left( \frac{1}{1} \tan^{-1} \left( \frac{x+2}{1} \right) \right) + C \]
\[ = \frac{1}{2} \log|x^2+4x+5| - \tan^{-1} (x+2) + C \]In simple words: We break down the top part of the fraction based on the derivative of the bottom part. This turns the problem into two easier integrals: one that becomes a logarithm and another that needs completing the square and then uses the tangent inverse formula.

🎯 Exam Tip: When the constant term in the denominator after completing the square is positive (like \( a^2 \)), the integral often involves an inverse tangent function. If it's negative (like \( -a^2 \)), it usually involves a logarithm.

Question 10. \( \frac{(3 \sin x - 2) \cos x}{5 - \cos^2 x - 4 \sin x} \)
Answer: We begin by simplifying the denominator using the identity \( \cos^2 x = 1 - \sin^2 x \). Then, we perform a substitution to simplify the integral into an algebraic form.
\[ \int \frac{(3 \sin x - 2) \cos x}{5 - \cos^2 x - 4 \sin x} \, dx \] Replace \( \cos^2 x \) with \( 1-\sin^2 x \):
\[ = \int \frac{(3 \sin x - 2) \cos x}{5 - (1-\sin^2 x) - 4 \sin x} \, dx \] \[ = \int \frac{(3 \sin x - 2) \cos x}{5 - 1 + \sin^2 x - 4 \sin x} \, dx \] \[ = \int \frac{(3 \sin x - 2) \cos x}{\sin^2 x - 4 \sin x + 4} \, dx \] Now, let \( t = \sin x \). Then \( dt = \cos x \, dx \). The integral transforms to:
\[ = \int \frac{3t-2}{t^2-4t+4} \, dt \] Notice that the denominator is a perfect square: \( t^2-4t+4 = (t-2)^2 \).
\[ = \int \frac{3t-2}{(t-2)^2} \, dt \] We can write \( 3t-2 \) as \( 3(t-2) + 4 \).
\[ = \int \frac{3(t-2)+4}{(t-2)^2} \, dt \] Split the fraction:
\[ = \int \left( \frac{3(t-2)}{(t-2)^2} + \frac{4}{(t-2)^2} \right) \, dt \] \[ = \int \left( \frac{3}{t-2} + 4(t-2)^{-2} \right) \, dt \] Now, integrate each term:
\[ = 3 \log|t-2| + 4 \frac{(t-2)^{-1}}{-1} + C \] \[ = 3 \log|t-2| - \frac{4}{t-2} + C \] Substitute back \( t = \sin x \). The value of \( \sin x \) is always between -1 and 1. So, \( t-2 \) will always be negative. This means we can write \( \log|t-2| \) as \( \log(2-t) \).
\[ = 3 \log| \sin x - 2 | - \frac{4}{\sin x - 2} + C \] \[ = 3 \log(2 - \sin x) + \frac{4}{2 - \sin x} + C \]In simple words: First, we change \( \cos^2 x \) to \( 1-\sin^2 x \). Then we let \( \sin x \) be \( t \), which makes the integral simpler. The bottom part becomes a perfect square. We split the fraction and integrate each part separately, finally putting \( \sin x \) back.

🎯 Exam Tip: Always look for trigonometric identities (like \( \sin^2 x + \cos^2 x = 1 \)) to simplify complex trigonometric integrals, especially those involving powers of \( \sin x \) and \( \cos x \).

Question 11. \( \frac{1}{2e^{2x} + 3e^x + 1} \)
Answer: This integral can be solved using a substitution involving \( e^x \).
Let \( t = e^x \). Then \( dt = e^x \, dx \), which means \( dx = \frac{dt}{e^x} = \frac{dt}{t} \). The integral becomes:
\[ \int \frac{1}{2(e^x)^2 + 3e^x + 1} \, dx = \int \frac{1}{2t^2 + 3t + 1} \cdot \frac{dt}{t} \] \[ = \int \frac{1}{t(2t^2 + 3t + 1)} \, dt \] The quadratic factor in the denominator can be factored: \( 2t^2+3t+1 = (2t+1)(t+1) \).
\[ = \int \frac{1}{t(t+1)(2t+1)} \, dt \] Now, we use partial fraction decomposition.
Let \( \frac{1}{t(t+1)(2t+1)} = \frac{A}{t} + \frac{B}{t+1} + \frac{C}{2t+1} \) Multiply by \( t(t+1)(2t+1) \):
\( 1 = A(t+1)(2t+1) + B t(2t+1) + C t(t+1) \) To find A, set \( t=0 \): \( 1 = A(1)(1) \implies A = 1 \) To find B, set \( t=-1 \): \( 1 = B(-1)(-2+1) \implies 1 = B(-1)(-1) \implies B = 1 \) To find C, set \( t=-\frac{1}{2} \): \( 1 = C(-\frac{1}{2})(-\frac{1}{2}+1) \implies 1 = C(-\frac{1}{2})(\frac{1}{2}) \implies 1 = -\frac{1}{4}C \implies C = -4 \) So, the integral becomes:
\[ \int \left( \frac{1}{t} + \frac{1}{t+1} - \frac{4}{2t+1} \right) \, dt \] Integrate each term:
\[ = \log|t| + \log|t+1| - 4 \frac{\log|2t+1|}{2} + C \] \[ = \log|t| + \log|t+1| - 2 \log|2t+1| + C \] Using logarithm properties: \( \log a + \log b - \log c = \log \left( \frac{ab}{c} \right) \).
\[ = \log \left| \frac{t(t+1)}{(2t+1)^2} \right| + C \] Finally, substitute back \( t = e^x \).
\[ = \log \left| \frac{e^x(e^x+1)}{(2e^x+1)^2} \right| + C \]In simple words: We replace \( e^x \) with \( t \) to simplify the problem. The bottom part of the fraction can be factored. Then, we use partial fractions to break the fraction into simpler parts that are easy to integrate using logarithms. Finally, we put \( e^x \) back into our answer.

🎯 Exam Tip: Integrals involving exponential terms like \( e^{2x} \) and \( e^x \) often benefit from the substitution \( t = e^x \), which transforms them into rational functions that can be solved using partial fractions.

Question 12. \( \frac{1}{\sqrt{4x^2-5x+1}} \)
Answer: For integrals with a square root of a quadratic in the denominator, we complete the square. First, we factor out the coefficient of \( x^2 \) from under the square root.
\[ \int \frac{1}{\sqrt{4x^2-5x+1}} \, dx \] \[ = \int \frac{1}{\sqrt{4(x^2-\frac{5}{4}x+\frac{1}{4})}} \, dx \] \[ = \int \frac{1}{2\sqrt{x^2-\frac{5}{4}x+\frac{1}{4}}} \, dx \] \[ = \frac{1}{2} \int \frac{1}{\sqrt{x^2-\frac{5}{4}x+\frac{1}{4}}} \, dx \] To complete the square for \( x^2-\frac{5}{4}x \), we add and subtract \( \left( \frac{5/4}{2} \right)^2 = \left( \frac{5}{8} \right)^2 = \frac{25}{64} \).
\[ = \frac{1}{2} \int \frac{1}{\sqrt{x^2-\frac{5}{4}x+\frac{25}{64}-\frac{25}{64}+\frac{1}{4}}} \, dx \] \[ = \frac{1}{2} \int \frac{1}{\sqrt{\left(x-\frac{5}{8}\right)^2 - \frac{25}{64} + \frac{16}{64}}} \, dx \] \[ = \frac{1}{2} \int \frac{1}{\sqrt{\left(x-\frac{5}{8}\right)^2 - \frac{9}{64}}} \, dx \] \[ = \frac{1}{2} \int \frac{1}{\sqrt{\left(x-\frac{5}{8}\right)^2 - \left(\frac{3}{8}\right)^2}} \, dx \] Using the formula \( \int \frac{1}{\sqrt{x^2-a^2}} \, dx = \log \left| x + \sqrt{x^2-a^2} \right| + C \), with \( x=\left(x-\frac{5}{8}\right) \) and \( a=\frac{3}{8} \).
\[ = \frac{1}{2} \log \left| \left(x-\frac{5}{8}\right) + \sqrt{\left(x-\frac{5}{8}\right)^2 - \left(\frac{3}{8}\right)^2} \right| + C \] Substitute the original quadratic back into the square root. The expression under the square root is equivalent to \( x^2-\frac{5}{4}x+\frac{1}{4} \).
\[ = \frac{1}{2} \log \left| \left(x-\frac{5}{8}\right) + \sqrt{x^2-\frac{5}{4}x+\frac{1}{4}} \right| + C \]In simple words: We take out the 4 from inside the square root. Then we make the expression inside the square root a perfect square minus a number. After that, we use a specific logarithm formula for integrals with square roots.

🎯 Exam Tip: When dealing with \( \sqrt{ax^2+bx+c} \), always factor out \( a \) first to get \( \sqrt{a}\sqrt{x^2+\frac{b}{a}x+\frac{c}{a}} \) before completing the square for the simpler quadratic.

Question 13. \( \frac{1}{\sqrt{5x-6-x^2}} \)
Answer: This integral involves a square root of a quadratic with a negative \( x^2 \) term. We complete the square inside the square root carefully.
\[ \int \frac{1}{\sqrt{5x-6-x^2}} \, dx \] Rearrange the terms inside the square root:
\[ = \int \frac{1}{\sqrt{-(x^2-5x+6)}} \, dx \] To complete the square for \( x^2-5x \), we add and subtract \( \left( \frac{-5}{2} \right)^2 = \frac{25}{4} \).
\[ = \int \frac{1}{\sqrt{-\left(x^2-5x+\frac{25}{4}-\frac{25}{4}+6\right)}} \, dx \] \[ = \int \frac{1}{\sqrt{-\left(\left(x-\frac{5}{2}\right)^2 - \frac{25}{4} + \frac{24}{4}\right)}} \, dx \] \[ = \int \frac{1}{\sqrt{-\left(\left(x-\frac{5}{2}\right)^2 - \frac{1}{4}\right)}} \, dx \] Distribute the negative sign:
\[ = \int \frac{1}{\sqrt{\frac{1}{4} - \left(x-\frac{5}{2}\right)^2}} \, dx \] \[ = \int \frac{1}{\sqrt{\left(\frac{1}{2}\right)^2 - \left(x-\frac{5}{2}\right)^2}} \, dx \] Using the formula \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + C \), with \( x=\left(x-\frac{5}{2}\right) \) and \( a=\frac{1}{2} \).
\[ = \sin^{-1} \left( \frac{x-\frac{5}{2}}{\frac{1}{2}} \right) + C \] \[ = \sin^{-1} \left( \frac{\frac{2x-5}{2}}{\frac{1}{2}} \right) + C \] \[ = \sin^{-1} (2x-5) + C \]In simple words: We rewrite the expression under the square root by taking out a minus sign and completing the square. This puts it into a form of a constant squared minus a variable term squared. Then, we use the inverse sine formula.

🎯 Exam Tip: When the \( x^2 \) term under the square root is negative, remember to factor out the negative sign first before completing the square to correctly apply the \( \sin^{-1} \) integral formula.

Question 14. \( \frac{1}{\sqrt{1-x-x^2}} \)
Answer: Similar to the previous question, this integral requires completing the square under the square root. We handle the negative \( x^2 \) term carefully.
\[ \int \frac{1}{\sqrt{1-x-x^2}} \, dx \] Rearrange and factor out the negative sign from the quadratic part:
\[ = \int \frac{1}{\sqrt{1-(x^2+x)}} \, dx \] To complete the square for \( x^2+x \), we add and subtract \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \).
\[ = \int \frac{1}{\sqrt{1-\left(x^2+x+\frac{1}{4}-\frac{1}{4}\right)}} \, dx \] \[ = \int \frac{1}{\sqrt{1-\left(\left(x+\frac{1}{2}\right)^2 - \frac{1}{4}\right)}} \, dx \] Distribute the negative sign:
\[ = \int \frac{1}{\sqrt{1-\left(x+\frac{1}{2}\right)^2 + \frac{1}{4}}} \, dx \] \[ = \int \frac{1}{\sqrt{\frac{5}{4} - \left(x+\frac{1}{2}\right)^2}} \, dx \] \[ = \int \frac{1}{\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - \left(x+\frac{1}{2}\right)^2}} \, dx \] Using the formula \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + C \), with \( x=\left(x+\frac{1}{2}\right) \) and \( a=\frac{\sqrt{5}}{2} \).
\[ = \sin^{-1} \left( \frac{x+\frac{1}{2}}{\frac{\sqrt{5}}{2}} \right) + C \] \[ = \sin^{-1} \left( \frac{\frac{2x+1}{2}}{\frac{\sqrt{5}}{2}} \right) + C \] \[ = \sin^{-1} \left( \frac{2x+1}{\sqrt{5}} \right) + C \]In simple words: We reorganize the terms under the square root, separating the constant and making the \( x \) terms into a perfect square. Because of the negative \( x^2 \) term, it leads to an inverse sine solution.

🎯 Exam Tip: Always double-check your arithmetic, especially when combining fractions (like \( 1 + \frac{1}{4} = \frac{5}{4} \)) during the completing-the-square process, as small errors can lead to incorrect \( a \) values.

Question 16. \( \frac{x+2}{x^2-2x+4} \)
Answer: We use the method of expressing the numerator in terms of the derivative of the denominator to simplify this integral.
Let \( x+2 = A \frac{d}{dx}(x^2-2x+4) + B \).
\( x+2 = A(2x-2) + B \)
\( x+2 = 2Ax - 2A + B \) Comparing coefficients of \( x \):
\( 2A = 1 \implies A = \frac{1}{2} \) Comparing constant terms:
\( -2A+B = 2 \)
\( -2\left(\frac{1}{2}\right) + B = 2 \)
\( -1+B = 2 \implies B = 3 \) So, \( x+2 = \frac{1}{2}(2x-2) + 3 \). Now, substitute this back into the integral:
\[ \int \frac{x+2}{x^2-2x+4} \, dx = \int \frac{\frac{1}{2}(2x-2) + 3}{x^2-2x+4} \, dx \] Split this into two integrals:
\[ = \frac{1}{2} \int \frac{2x-2}{x^2-2x+4} \, dx + 3 \int \frac{1}{x^2-2x+4} \, dx \] For the first integral, let \( u = x^2-2x+4 \), then \( du = (2x-2) \, dx \). So, \( \int \frac{du}{u} = \log|u| \).
\[ = \frac{1}{2} \log|x^2-2x+4| + 3 \int \frac{1}{x^2-2x+4} \, dx \] For the second integral, we complete the square in the denominator: \( x^2-2x+4 = x^2-2x+1+3 = (x-1)^2+(\sqrt{3})^2 \).
\[ = \frac{1}{2} \log|x^2-2x+4| + 3 \int \frac{1}{(x-1)^2+(\sqrt{3})^2} \, dx \] Using the formula \( \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), with \( x=(x-1) \) and \( a=\sqrt{3} \).
\[ = \frac{1}{2} \log|x^2-2x+4| + 3 \left( \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x-1}{\sqrt{3}} \right) \right) + C \]
\[ = \frac{1}{2} \log|x^2-2x+4| + \sqrt{3} \tan^{-1} \left( \frac{x-1}{\sqrt{3}} \right) + C \]In simple words: We rewrite the numerator to match the derivative of the denominator, which helps us divide the integral into two parts. One part becomes a logarithm. For the other part, we complete the square in the bottom and use the inverse tangent formula.

🎯 Exam Tip: Always remember that the derivative of a quadratic \( ax^2+bx+c \) is \( 2ax+b \). This helps set up the constants A and B correctly when decomposing the numerator.

Question 17. \( \frac{x+1}{\sqrt{x^2-x+1}} \)
Answer: The provided solution addresses the integral \( \int \frac{x+2}{\sqrt{x^2-2x+4}} dx \) by splitting the numerator into two parts. This makes it easier to integrate. The first part becomes a direct derivative of the denominator, and the second part uses a standard integral form after completing the square in the denominator.
To evaluate \( \int \frac{x+2}{\sqrt{x^2-2x+4}} dx \):
We can rewrite the numerator \( x+2 \) using the derivative of the quadratic expression in the denominator \( x^2-2x+4 \). The derivative of \( x^2-2x+4 \) is \( 2x-2 \).
So, \( x+2 = \frac{1}{2}(2x-2) + 3 \)
Substituting this back into the integral:
\( \int \frac{\frac{1}{2}(2x-2) + 3}{\sqrt{x^2-2x+4}} dx \)
This can be split into two separate integrals:
\( = \frac{1}{2} \int \frac{2x-2}{\sqrt{x^2-2x+4}} dx + 3 \int \frac{1}{\sqrt{x^2-2x+4}} dx \)
For the first integral, let \( t = x^2-2x+4 \). Then, \( dt = (2x-2)dx \).
\( \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} = t^{1/2} = \sqrt{x^2-2x+4} \)
For the second integral, we complete the square in the denominator:
\( x^2-2x+4 = (x^2-2x+1) + 3 = (x-1)^2 + (\sqrt{3})^2 \)
So, \( 3 \int \frac{1}{\sqrt{(x-1)^2 + (\sqrt{3})^2}} dx \)
This is a standard integral form \( \int \frac{1}{\sqrt{y^2+a^2}} dy = \log |y + \sqrt{y^2+a^2}| + C \).
\( = 3 \log \left| (x-1) + \sqrt{(x-1)^2 + (\sqrt{3})^2} \right| + C \)
\( = 3 \log \left| (x-1) + \sqrt{x^2-2x+4} \right| + C \)
Combining both parts, the final solution for the integral \( \int \frac{x+2}{\sqrt{x^2-2x+4}} dx \) is:
\( \sqrt{x^2-2x+4} + 3 \log \left| (x-1) + \sqrt{x^2-2x+4} \right| + C \)In simple words: To solve this integral, we break it into two parts. One part is easy because the top is the derivative of the bottom. For the other part, we complete the square under the root and use a special log formula.

🎯 Exam Tip: When integrating fractions with a linear term in the numerator and a quadratic root in the denominator, always try to split the numerator into a derivative part and a constant part. This makes the integration simpler and more manageable.

Question 18. \( \frac{3x+3}{\sqrt{x^2+2x+2}} \)
Answer: To solve this integral, we first rewrite the numerator in terms of the derivative of the quadratic expression under the square root in the denominator. This allows us to split the integral into two simpler forms, which are then evaluated separately using standard integration formulas.
The integral is \( \int \frac{x+3}{\sqrt{x^2+2x+2}} dx \).
Let the numerator \( x+3 \) be expressed as \( A(2x+2) + B \), where \( 2x+2 \) is the derivative of \( x^2+2x+2 \).
\( x+3 = A(2x+2) + B \)
Comparing the coefficients of \( x \): \( 1 = 2A \implies A = \frac{1}{2} \)
Comparing the constant terms: \( 3 = 2A + B \implies 3 = 2\left(\frac{1}{2}\right) + B \implies 3 = 1 + B \implies B = 2 \)
So, the numerator \( x+3 \) can be written as \( \frac{1}{2}(2x+2) + 2 \).
Substitute this into the integral:
\( \int \frac{\frac{1}{2}(2x+2) + 2}{\sqrt{x^2+2x+2}} dx \)
Split the integral into two parts:
\( = \frac{1}{2} \int \frac{2x+2}{\sqrt{x^2+2x+2}} dx + 2 \int \frac{1}{\sqrt{x^2+2x+2}} dx \)
For the first integral, let \( t = x^2+2x+2 \). Then, \( dt = (2x+2)dx \).
\( \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} = \sqrt{t} = \sqrt{x^2+2x+2} \)
For the second integral, complete the square for the quadratic expression in the denominator:
\( x^2+2x+2 = (x^2+2x+1) + 1 = (x+1)^2 + 1^2 \)
So, \( 2 \int \frac{1}{\sqrt{(x+1)^2+1^2}} dx \)
This is a standard integral form \( \int \frac{1}{\sqrt{y^2+a^2}} dy = \log |y + \sqrt{y^2+a^2}| + C \).
\( = 2 \log \left| (x+1) + \sqrt{(x+1)^2+1} \right| + C \)
\( = 2 \log \left| (x+1) + \sqrt{x^2+2x+2} \right| + C \)
Combining both parts, the final solution is:
\( \sqrt{x^2+2x+2} + 2 \log \left| (x+1) + \sqrt{x^2+2x+2} \right| + C \)In simple words: We break the top part of the fraction into two pieces. One piece helps us solve the integral directly. For the second piece, we adjust the bottom part by completing the square and use a special logarithm rule for square roots.

🎯 Exam Tip: Ensure careful algebraic manipulation when splitting the numerator and completing the square. A small error in these steps can lead to an incorrect final result.

Question 19. \( \sqrt{\sec x - 1} \) dx
Answer: The provided solution involves a series of transformations and substitutions to evaluate an integral. It starts by attempting to use trigonometric identities, then applies a variable substitution. However, there's a shift in the integral form during the steps, leading to the solution of a different integral after the substitution. The steps show the application of specific integration formulas based on the transformed form.
To integrate \( \sqrt{\sec x - 1} \):
The provided solution begins with:
\( \int \frac{\cos x}{1+\cos x} dx \)
Then it proceeds to:
\( = \int \frac{1-\cos^2 x}{\cos x (1+\cos x)} dx \)
A substitution is then introduced:
Let \( \cos x = t \).
Then \( -\sin x dx = dt \).
The integral is shown as:
\( = - \int \frac{1}{t(1+t)} dx \) (Note: The variable of integration in this step is incorrectly written as `dx` instead of `dt`.)
Following this, the solution abruptly changes to integrate a different form:
\( = - \int \frac{1}{\sqrt{t^2+t+\frac{1}{4}-\frac{1}{4}}} dt \) (This integral form is distinct from the one derived in the previous step.)
Complete the square for the expression under the square root:
\( = - \int \frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} dt \)
This integral is of the form \( \int \frac{1}{\sqrt{y^2-a^2}} dy = \log |y + \sqrt{y^2-a^2}| + C \).
\( = - \log \left| t + \frac{1}{2} + \sqrt{\left(t+\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2} \right| + C \)
Simplifying the term under the square root:
\( = - \log \left| t + \frac{1}{2} + \sqrt{t^2+t} \right| + C \)
Finally, substituting back \( t = \cos x \):
\( = - \log \left| \cos x + \frac{1}{2} + \sqrt{\cos^2 x + \cos x} \right| + C \)In simple words: This solution takes several turns. It tries to change the problem using math rules for angles, then switches letters in a substitution. After that, the problem itself changes, and a special rule for square roots is used to find the answer. Finally, the original letter is put back.

🎯 Exam Tip: When solving complex trigonometric integrals, look for ways to simplify the expression using identities before applying substitutions. Always double-check the variable of integration to avoid errors.

Question 20. \( \frac{\sin (x-\alpha)}{\sqrt{\sin (x+\alpha)}} \)
Answer: The solution addresses an integral with trigonometric functions by first expanding the numerator using an angle subtraction formula. It then splits the integral into two parts, \( I_1 \) and \( I_2 \), each solved using an appropriate substitution and standard integral formulas related to inverse sine and logarithms, respectively. The final result combines the solutions of these two parts.
The integral is \( \int \frac{\sin (x-\alpha)}{\sqrt{\sin (x+\alpha)}} dx \).
The solution proceeds by using \( \sqrt{\sin^2 x - \sin^2 \alpha} \) in the denominator for calculation, assuming this is the intended form of the problem after simplification or correction.
First, expand the numerator using the angle subtraction identity \( \sin(A-B) = \sin A \cos B - \cos A \sin B \):
\( \sin(x-\alpha) = \sin x \cos \alpha - \cos x \sin \alpha \)
The integral then becomes:
\( \int \frac{\sin x \cos \alpha - \cos x \sin \alpha}{\sqrt{\sin^2 x - \sin^2 \alpha}} dx \)
This integral is split into two distinct parts, \( I_1 \) and \( I_2 \):
\( I_1 = \int \frac{\sin x \cos \alpha}{\sqrt{\sin^2 x - \sin^2 \alpha}} dx \)
Rewrite \( \sin^2 x \) as \( 1-\cos^2 x \) in the denominator:
\( I_1 = \cos \alpha \int \frac{\sin x}{\sqrt{1-\cos^2 x - \sin^2 \alpha}} dx = \cos \alpha \int \frac{\sin x}{\sqrt{\cos^2 \alpha - \cos^2 x}} dx \)
Let \( t = \cos x \), then \( dt = -\sin x dx \).
\( I_1 = \cos \alpha \int \frac{-dt}{\sqrt{\cos^2 \alpha - t^2}} = -\cos \alpha \int \frac{1}{\sqrt{(\cos \alpha)^2 - t^2}} dt \)
This is a standard integral form \( \int \frac{1}{\sqrt{a^2-y^2}} dy = \sin^{-1}(\frac{y}{a}) + C \).
\( I_1 = -\cos \alpha \sin^{-1} \left( \frac{t}{\cos \alpha} \right) = -\cos \alpha \sin^{-1} \left( \frac{\cos x}{\cos \alpha} \right) \)
Now for the second part, \( I_2 \):
\( I_2 = \int \frac{\cos x \sin \alpha}{\sqrt{\sin^2 x - \sin^2 \alpha}} dx \)
Let \( y = \sin x \), then \( dy = \cos x dx \).
\( I_2 = \sin \alpha \int \frac{dy}{\sqrt{y^2 - \sin^2 \alpha}} \)
This is a standard integral form \( \int \frac{1}{\sqrt{y^2-a^2}} dy = \log |y + \sqrt{y^2-a^2}| + C \).
\( I_2 = \sin \alpha \log \left| y + \sqrt{y^2 - \sin^2 \alpha} \right| \)
Substituting back \( y = \sin x \):
\( I_2 = \sin \alpha \log \left| \sin x + \sqrt{\sin^2 x - \sin^2 \alpha} \right| \)
Combining the results for \( I_1 \) and \( I_2 \), the final integral \( I \) is:
\( I = I_1 - I_2 = -\cos \alpha \sin^{-1} \left( \frac{\cos x}{\cos \alpha} \right) - \sin \alpha \log \left| \sin x + \sqrt{\sin^2 x - \sin^2 \alpha} \right| + C \)In simple words: We first break down the top part of the fraction using a rule for sines. Then, we split the whole problem into two smaller problems. We use different substitutions and special math rules to solve each part, and then put them back together for the final answer.

🎯 Exam Tip: Remember to apply trigonometric identities correctly to simplify complex expressions. Always check if the denominator can be expressed in a standard form, such as \( \sqrt{a^2-x^2} \) or \( \sqrt{x^2-a^2} \), as this guides the choice of integral formula.

Question 22. \( \frac{e^x}{e^{2x} + 6e^x + 5} \)
Answer: The provided solution demonstrates how to integrate a rational function where the degree of the numerator is greater than or equal to the degree of the denominator. This is achieved by first using polynomial long division to simplify the expression into a polynomial part and a proper rational function part. The polynomial terms are integrated directly, and the remaining rational function is integrated by completing the square in the denominator and applying a standard formula.
The solution provided is for the integral \( \int \frac{x^3}{x^2+x+1} dx \).
First, perform polynomial long division on \( \frac{x^3}{x^2+x+1} \):
\( x^3 = x(x^2+x+1) - x^2 - x \)
\( = x(x^2+x+1) - (x^2+x+1) + 1 \)
So, \( \frac{x^3}{x^2+x+1} = x - 1 + \frac{1}{x^2+x+1} \)
Now, integrate each term:
\( \int \left( x - 1 + \frac{1}{x^2+x+1} \right) dx \)
\( = \int x dx - \int 1 dx + \int \frac{1}{x^2+x+1} dx \)
\( = \frac{x^2}{2} - x + \int \frac{1}{x^2+x+1} dx \)
For the remaining integral, \( \int \frac{1}{x^2+x+1} dx \), complete the square in the denominator:
\( x^2+x+1 = \left( x^2+x+\frac{1}{4} \right) + 1 - \frac{1}{4} = \left( x + \frac{1}{2} \right)^2 + \frac{3}{4} \)
We can write \( \frac{3}{4} \) as \( \left( \frac{\sqrt{3}}{2} \right)^2 \).
So the integral becomes: \( \int \frac{1}{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} dx \)
This is of the standard integral form \( \int \frac{1}{y^2+a^2} dy = \frac{1}{a} \tan^{-1} \left( \frac{y}{a} \right) + C \).
Here, \( y = x+\frac{1}{2} \) and \( a = \frac{\sqrt{3}}{2} \).
\( = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1} \left( \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C \)
\( = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{\frac{2x+1}{2}}{\frac{\sqrt{3}}{2}} \right) + C \)
\( = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2x+1}{\sqrt{3}} \right) + C \)
Combining all parts, the final solution for the integral \( \int \frac{x^3}{x^2+x+1} dx \) is:
\( \frac{x^2}{2} - x + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2x+1}{\sqrt{3}} \right) + C \)In simple words: When the top part of the fraction is of a higher power than the bottom, we first divide them like we do with numbers. This gives us some simple parts and a new, simpler fraction. We then integrate each part separately, using a special tangent rule for the remaining fraction.

🎯 Exam Tip: Always perform polynomial long division when the degree of the numerator is greater than or equal to the degree of the denominator in rational function integrals. This simplifies the problem into integrable forms.