GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.2

Question 1. Integrate \( \frac{2 x}{1+x^{2}} \)
Answer: Let the integral be \( I = \int \frac{2 x}{1+x^{2}} dx \). We can make a substitution here. Put \( 1 + x^{2} = t \). Then, by differentiation, \( 2x \, dx = dt \).
\( \implies I = \int \frac{dt}{t} \)
\( \implies I = \log |t| + C \)
\( \implies I = \log |1+x^{2}| + C \). Since \( 1+x^2 \) is always positive, the absolute value can be removed.
\( \implies I = \log (1+x^{2}) + C \).
In simple words: We find the integral by letting the denominator be 't' and then replacing '2x dx' with 'dt'. After integrating, we substitute 't' back with its original expression.

Exam Tip: For rational functions, always check if the numerator is the derivative of the denominator, as this suggests a direct substitution method using \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \).

Question 2. Integrate \( \frac{(\log x)^{2}}{x} \)
Answer: Let the integral be \( I = \int \frac{(\log x)^{2}}{x} dx \). We should use a substitution method. Put \( \log x = t \). Then, differentiating, \( \frac{1}{x} dx = dt \).
\( \implies I = \int t^{2} dt \)
\( \implies I = \frac{t^{3}}{3} + C \) Now, substitute back for \( t \).
\( \implies I = \frac{(\log |x|)^{3}}{3} + C \).
In simple words: To solve this, we let 'log x' be 't'. Then, '1/x dx' becomes 'dt'. After that, we integrate 't squared' and put 'log x' back in place of 't' for the final answer.

Exam Tip: When you see a function and its derivative multiplied together, consider substituting the function to simplify the integral. For logarithms, \( \frac{1}{x} \) is a clear hint for substitution.

Question 3. Integrate \( \frac{1}{x+x \log x} \)
Answer: Let the integral be \( I = \int \frac{1}{x+x \log x} dx \). First, we factor out \( x \) from the denominator.
\( \implies I = \int \frac{1}{x(1+\log x)} dx \) Now, we can make a substitution. Put \( 1 + \log x = t \). Differentiating both sides gives \( \frac{1}{x} dx = dt \).
\( \implies I = \int \frac{1}{t} dt \)
\( \implies I = \log |t| + C \) Substitute back \( t = 1 + \log x \).
\( \implies I = \log |1 + \log x| + C \).
In simple words: We first take 'x' out of the bottom part of the fraction. Then, we let '1 + log x' be 't' and change '1/x dx' to 'dt'. Finally, we integrate '1 over t' and put '1 + log x' back in.

Exam Tip: Always try to factorize the denominator first. This can often reveal a suitable substitution if one of the factors is related to the derivative of another part of the expression.

Question 4. Integrate \( \sin x \sin(\cos x) \)
Answer: Let the integral be \( I = \int \sin x \sin(\cos x) dx \). We should perform a substitution here. Put \( \cos x = t \). Differentiating both sides, we get \( -\sin x \, dx = dt \), which means \( \sin x \, dx = -dt \).
\( \implies I = \int \sin(t) (-dt) \)
\( \implies I = - \int \sin t \, dt \) The integral of \( \sin t \) is \( -\cos t \).
\( \implies I = - (-\cos t) + C \)
\( \implies I = \cos t + C \) Now, substitute back \( t = \cos x \).
\( \implies I = \cos (\cos x) + C \).
In simple words: We solve this by making 'cos x' equal to 't'. This changes 'sin x dx' to '-dt'. We then integrate '-sin t', which gives 'cos t'. After that, we replace 't' with 'cos x' to get the final answer.

Exam Tip: When an integral involves a composite function (like \( \sin(\cos x) \)), consider substituting the inner function. Also, be careful with the signs when differentiating for the substitution.

Question 5. Integrate \( \sin(ax + b)\cos(ax + b) \)
Answer: Let the integral be \( I = \int \sin(ax + b)\cos(ax + b)dx \). We can use the identity \( \sin(2\theta) = 2\sin\theta\cos\theta \).
\( \implies I = \int \frac{1}{2} (2\sin(ax+b)\cos(ax+b))dx \)
\( \implies I = \frac{1}{2} \int \sin(2(ax+b))dx \)
\( \implies I = \frac{1}{2} \int \sin(2ax+2b)dx \) Now, perform a substitution. Put \( 2ax + 2b = t \). Differentiating both sides gives \( 2a \, dx = dt \), so \( dx = \frac{1}{2a} dt \).
\( \implies I = \frac{1}{2} \int \sin t \left(\frac{1}{2a}\right) dt \)
\( \implies I = \frac{1}{4a} \int \sin t \, dt \) The integral of \( \sin t \) is \( -\cos t \).
\( \implies I = \frac{1}{4a} (-\cos t) + C \)
\( \implies I = - \frac{1}{4a} \cos(2ax+2b) + C \).
In simple words: We can use the double-angle formula for sine to simplify the problem. After that, we replace '2ax + 2b' with 't' and integrate 'sin t'. Finally, we put the original expression back in place of 't'.

Exam Tip: Recognize trigonometric identities that simplify products into sums or single terms, especially when dealing with sine and cosine functions of the same argument. This often makes integration much easier.

Question 6. Integrate \( \sqrt{ax+b} \)
Answer: Let the integral be \( I = \int \sqrt{ax+b} \, dx \). We can rewrite the square root as a power.
\( \implies I = \int (ax+b)^{1/2} dx \) Now, perform a substitution. Put \( ax + b = t \). Differentiating both sides gives \( a \, dx = dt \), so \( dx = \frac{1}{a} dt \).
\( \implies I = \int t^{1/2} \left(\frac{1}{a}\right) dt \)
\( \implies I = \frac{1}{a} \int t^{1/2} dt \) Using the power rule for integration \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \).
\( \implies I = \frac{1}{a} \left( \frac{t^{1/2 + 1}}{1/2 + 1} \right) + C \)
\( \implies I = \frac{1}{a} \left( \frac{t^{3/2}}{3/2} \right) + C \)
\( \implies I = \frac{1}{a} \left( \frac{2}{3} t^{3/2} \right) + C \)
\( \implies I = \frac{2}{3a} (ax+b)^{3/2} + C \).
In simple words: We change the square root to a power of 1/2. Then, we let 'ax + b' be 't'. After that, we integrate 't to the power of 1/2' and put 'ax + b' back in place of 't' for the final answer.

Exam Tip: For functions of the form \( (ax+b)^n \), use the substitution \( u = ax+b \). Remember that \( dx = \frac{1}{a} du \), and you'll need to multiply by \( \frac{1}{a} \) after integrating with respect to \( u \).

Question 7. Integrate \( x\sqrt{x+2} \)
Answer: Let the integral be \( I = \int x\sqrt{x+2}dx \). To make this easier, we can rewrite \( x \) in terms of \( (x+2) \).
\( \implies I = \int [(x+2) - 2]\sqrt{x+2}dx \) Distribute \( \sqrt{x+2} \) into the bracket.
\( \implies I = \int (x+2)\sqrt{x+2}dx - \int 2\sqrt{x+2}dx \)
\( \implies I = \int (x+2)^{1} (x+2)^{1/2}dx - 2 \int (x+2)^{1/2}dx \)
\( \implies I = \int (x+2)^{3/2}dx - 2 \int (x+2)^{1/2}dx \) Now, we can integrate each term. Using the formula \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C \). Here, \( a=1 \). For the first term: \( \frac{(x+2)^{3/2 + 1}}{3/2 + 1} = \frac{(x+2)^{5/2}}{5/2} = \frac{2}{5}(x+2)^{5/2} \). For the second term: \( -2 \frac{(x+2)^{1/2 + 1}}{1/2 + 1} = -2 \frac{(x+2)^{3/2}}{3/2} = -2 \cdot \frac{2}{3}(x+2)^{3/2} = - \frac{4}{3}(x+2)^{3/2} \). Combining these, we get:
\( \implies I = \frac{2}{5}(x+2)^{5/2} - \frac{4}{3}(x+2)^{3/2} + C \).
In simple words: We change 'x' to 'x+2-2' to help simplify the problem. Then, we multiply 'sqrt(x+2)' by both parts. This lets us integrate each part separately using the power rule, giving us the final solution.

Exam Tip: When integrating \( x\sqrt{ax+b} \) or similar forms, try to express \( x \) in terms of \( (ax+b) \) (e.g., \( x = \frac{1}{a}(ax+b - b) \)) to make the substitution simpler and allow for direct power rule integration after expansion.

Question 8. Integrate \( x\sqrt{1+2 x^{2}} \)
Answer: Let the integral be \( I = \int x\sqrt{1+2x^{2}}dx \). We should use a substitution here. Put \( 1 + 2x^{2} = t \). Differentiating both sides gives \( 4x \, dx = dt \), so \( x \, dx = \frac{1}{4} dt \).
\( \implies I = \int \sqrt{t} \left(\frac{1}{4}\right) dt \)
\( \implies I = \frac{1}{4} \int t^{1/2} dt \) Using the power rule for integration:
\( \implies I = \frac{1}{4} \left( \frac{t^{1/2 + 1}}{1/2 + 1} \right) + C \)
\( \implies I = \frac{1}{4} \left( \frac{t^{3/2}}{3/2} \right) + C \)
\( \implies I = \frac{1}{4} \left( \frac{2}{3} t^{3/2} \right) + C \)
\( \implies I = \frac{1}{6} t^{3/2} + C \) Substitute back \( t = 1+2x^{2} \).
\( \implies I = \frac{1}{6} (1+2x^{2})^{3/2} + C \).
In simple words: We make '1 + 2x squared' equal to 't'. Then 'x dx' becomes '1/4 dt'. We integrate 'sqrt t' and then replace 't' with '1 + 2x squared' to get the final answer.

Exam Tip: If you see a term like \( x^n \) and its derivative \( x^{n-1} \) (or a multiple of it) in the integral, substitution is often the key. Here, \( x \) is a part of the derivative of \( 1+2x^2 \).

Question 9. Integrate \( (4x + 2)\sqrt{x^{2}+x+1} \)
Answer: Let the integral be \( I = \int (4x + 2)\sqrt{x^{2}+x+1}dx \). First, factor out 2 from \( (4x + 2) \).
\( \implies I = \int 2(2x + 1)\sqrt{x^{2}+x+1}dx \)
\( \implies I = 2 \int (2x + 1)\sqrt{x^{2}+x+1}dx \) Now, perform a substitution. Put \( x^{2}+x+1 = t \). Differentiating both sides gives \( (2x+1) \, dx = dt \).
\( \implies I = 2 \int \sqrt{t} \, dt \)
\( \implies I = 2 \int t^{1/2} dt \) Using the power rule for integration:
\( \implies I = 2 \left( \frac{t^{1/2+1}}{1/2+1} \right) + C \)
\( \implies I = 2 \left( \frac{t^{3/2}}{3/2} \right) + C \)
\( \implies I = 2 \left( \frac{2}{3} t^{3/2} \right) + C \)
\( \implies I = \frac{4}{3} t^{3/2} + C \) Substitute back \( t = x^{2}+x+1 \).
\( \implies I = \frac{4}{3} (x^{2}+x+1)^{3/2} + C \).
In simple words: We first take out the common factor '2'. Then, we let 'x squared + x + 1' be 't'. This makes '(2x + 1) dx' become 'dt'. We then integrate '2 times sqrt t' and put the original expression back for 't' in the solution.

Exam Tip: Always look for common factors or rearrangements that make the expression simpler. If one part of the integrand is the derivative of another part, a substitution is usually effective.

Question 10. Integrate \( \frac{1}{x-\sqrt{x}} \)
Answer: Let the integral be \( I = \int \frac{1}{x-\sqrt{x}} dx \). First, factor out \( \sqrt{x} \) from the denominator.
\( \implies I = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} dx \) Now, perform a substitution. Put \( \sqrt{x}-1 = t \). Differentiating both sides, we get \( \frac{1}{2\sqrt{x}} dx = dt \). This means \( \frac{1}{\sqrt{x}} dx = 2dt \).
\( \implies I = \int \frac{1}{t} (2dt) \)
\( \implies I = 2 \int \frac{1}{t} dt \)
\( \implies I = 2 \log |t| + C \) Substitute back \( t = \sqrt{x}-1 \).
\( \implies I = 2 \log |\sqrt{x}-1| + C \).
In simple words: We start by taking 'sqrt x' out of the bottom of the fraction. Then, we let 'sqrt x - 1' be 't'. This changes '1/sqrt x dx' to '2dt'. We then integrate '2 over t' and replace 't' with 'sqrt x - 1' to get the final answer.

Exam Tip: When dealing with expressions involving \( x \) and \( \sqrt{x} \), factoring \( \sqrt{x} \) is often a good first step. This can help reveal a suitable substitution that simplifies the integral.

Question 11. Integrate \( \frac{x}{\sqrt{x+4}} \)
Answer: Let the integral be \( I = \int \frac{x}{\sqrt{x+4}} dx \). To simplify, we can write \( x \) in terms of \( (x+4) \).
\( \implies I = \int \frac{(x+4)-4}{\sqrt{x+4}} dx \) Split the fraction into two terms.
\( \implies I = \int \left( \frac{x+4}{\sqrt{x+4}} - \frac{4}{\sqrt{x+4}} \right) dx \)
\( \implies I = \int (x+4)^{1/2} dx - 4 \int (x+4)^{-1/2} dx \) Now, integrate each term using the power rule \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C \). Here, \( a=1 \). For the first term: \( \frac{(x+4)^{1/2+1}}{1/2+1} = \frac{(x+4)^{3/2}}{3/2} = \frac{2}{3}(x+4)^{3/2} \). For the second term: \( -4 \frac{(x+4)^{-1/2+1}}{-1/2+1} = -4 \frac{(x+4)^{1/2}}{1/2} = -4 \cdot 2(x+4)^{1/2} = -8(x+4)^{1/2} \). Combining these, we get:
\( \implies I = \frac{2}{3}(x+4)^{3/2} - 8(x+4)^{1/2} + C \) We can factor out \( \frac{2}{3}(x+4)^{1/2} \).
\( \implies I = \frac{2}{3}(x+4)^{1/2} [(x+4) - 12] + C \)
\( \implies I = \frac{2}{3}\sqrt{x+4}(x-8) + C \).
In simple words: We rewrite 'x' as 'x+4-4' to simplify the problem. Then we break the fraction into two parts and integrate each part using the power rule. Finally, we combine and simplify to get the answer.

Exam Tip: When the numerator has \( x \) and the denominator has \( \sqrt{ax+b} \), try to express the numerator in terms of \( (ax+b) \) to make the integration more straightforward. This allows you to split the integral into simpler power functions.

Question 12. Integrate \( (x^3-1)^{1/3}x^5 \)
Answer: Let the integral be \( I = \int (x^{3}-1)^{1/3}x^{5} dx \). We can rewrite \( x^{5} \) as \( x^{3} \cdot x^{2} \).
\( \implies I = \int (x^{3}-1)^{1/3}x^{3} \cdot x^{2} dx \) Now, perform a substitution. Put \( x^{3} = t \). Differentiating both sides gives \( 3x^{2} dx = dt \), so \( x^{2} dx = \frac{1}{3} dt \). Also, from \( x^3 = t \), we can say \( x^3-1 = t-1 \).
\( \implies I = \int (t-1)^{1/3} t \left(\frac{1}{3} dt\right) \)
\( \implies I = \frac{1}{3} \int t(t-1)^{1/3} dt \) Distribute \( t \) into the bracket.
\( \implies I = \frac{1}{3} \int [(t-1)+1](t-1)^{1/3} dt \)
\( \implies I = \frac{1}{3} \int [(t-1)^{4/3} + (t-1)^{1/3}] dt \) Now, integrate each term using the power rule \( \int u^n du = \frac{u^{n+1}}{n+1} + C \).
\( \implies I = \frac{1}{3} \left[ \frac{(t-1)^{4/3+1}}{4/3+1} + \frac{(t-1)^{1/3+1}}{1/3+1} \right] + C \)
\( \implies I = \frac{1}{3} \left[ \frac{(t-1)^{7/3}}{7/3} + \frac{(t-1)^{4/3}}{4/3} \right] + C \)
\( \implies I = \frac{1}{3} \left[ \frac{3}{7} (t-1)^{7/3} + \frac{3}{4} (t-1)^{4/3} \right] + C \)
\( \implies I = \frac{1}{7} (t-1)^{7/3} + \frac{1}{4} (t-1)^{4/3} + C \) Substitute back \( t = x^{3} \).
\( \implies I = \frac{1}{7} (x^{3}-1)^{7/3} + \frac{1}{4} (x^{3}-1)^{4/3} + C \).
In simple words: We rewrite 'x to the power of 5' as 'x cubed times x squared'. Then, we let 'x cubed' be 't'. This makes 'x squared dx' become '1/3 dt'. We then integrate the new expression by separating it and using the power rule, finally putting 'x cubed' back for 't'.

Exam Tip: When you have a product like \( (f(x))^n \cdot g(x) \), and \( g(x) \) can be related to \( f'(x) \), substitution is likely the best approach. If \( x^3 \) is part of the base, and \( x^2 \) is present elsewhere, consider \( t=x^3 \).

Question 13. Integrate \( \frac{x^{2}}{\left(2+3 x^{3}\right)^{3}} \)
Answer: Let the integral be \( I = \int \frac{x^{2}}{(2+3x^{3})^{3}} dx \). We should use a substitution here. Put \( 2 + 3x^{3} = t \). Differentiating both sides gives \( 9x^{2} dx = dt \), so \( x^{2} dx = \frac{1}{9} dt \).
\( \implies I = \int \frac{1}{t^{3}} \left(\frac{1}{9} dt\right) \)
\( \implies I = \frac{1}{9} \int t^{-3} dt \) Using the power rule for integration:
\( \implies I = \frac{1}{9} \left( \frac{t^{-3+1}}{-3+1} \right) + C \)
\( \implies I = \frac{1}{9} \left( \frac{t^{-2}}{-2} \right) + C \)
\( \implies I = \frac{1}{9} \left( -\frac{1}{2t^{2}} \right) + C \)
\( \implies I = - \frac{1}{18t^{2}} + C \) Substitute back \( t = 2+3x^{3} \).
\( \implies I = - \frac{1}{18(2+3x^{3})^{2}} + C \).
In simple words: We let '2 + 3x cubed' be 't'. This means 'x squared dx' becomes '1/9 dt'. We then integrate 't to the power of -3' and replace 't' with '2 + 3x cubed' to find the final answer.

Exam Tip: For expressions involving \( (ax^n+b)^m \cdot x^{n-1} \), a substitution of \( t = ax^n+b \) is usually effective. Ensure you correctly manage the constant factor from the differentiation.

Question 14. Integrate \( \frac{1}{x(\log x)^{m}} \)
Answer: Let the integral be \( I = \int \frac{1}{x(\log x)^{m}} dx \), where \( x > 0 \). We should use a substitution here. Put \( \log x = t \). Differentiating both sides gives \( \frac{1}{x} dx = dt \).
\( \implies I = \int \frac{1}{t^{m}} dt \)
\( \implies I = \int t^{-m} dt \) Using the power rule for integration \( \int t^n dt = \frac{t^{n+1}}{n+1} + C \) (where \( n \neq -1 \)).
\( \implies I = \frac{t^{-m+1}}{-m+1} + C \) Substitute back \( t = \log x \).
\( \implies I = \frac{(\log x)^{1-m}}{1-m} + C \). Note: If \( m=1 \), the integral would be \( \int \frac{1}{t} dt = \log|t| + C = \log|\log x| + C \).
In simple words: We let 'log x' be 't', so '1/x dx' becomes 'dt'. Then we integrate 't to the power of -m'. Finally, we substitute 'log x' back in place of 't' to get the solution.

Exam Tip: When you see \( \log x \) and \( \frac{1}{x} \) together in an integral, it's a strong hint to substitute \( t = \log x \). Remember the special case for \( m=1 \) where the integral becomes a logarithm itself.

Question 15. Integrate \( \frac{x}{9-4 x^{2}} \)
Answer: Let the integral be \( I = \int \frac{x}{9-4x^{2}} dx \). We should use a substitution here. Put \( 9 - 4x^{2} = t \). Differentiating both sides gives \( -8x \, dx = dt \), so \( x \, dx = - \frac{1}{8} dt \).
\( \implies I = \int \frac{1}{t} \left(-\frac{1}{8} dt\right) \)
\( \implies I = - \frac{1}{8} \int \frac{1}{t} dt \) The integral of \( \frac{1}{t} \) is \( \log |t| \).
\( \implies I = - \frac{1}{8} \log |t| + C \) Substitute back \( t = 9-4x^{2} \).
\( \implies I = - \frac{1}{8} \log |9-4x^{2}| + C \).
In simple words: We let '9 - 4x squared' be 't'. This changes 'x dx' to '-1/8 dt'. We then integrate '-1/8 times 1 over t' and replace 't' with '9 - 4x squared' to find the answer.

Exam Tip: For rational functions, if the numerator is a multiple of the derivative of the denominator, a simple substitution \( t = \text{denominator} \) will simplify the integral into the form \( \int \frac{1}{t} dt \).

Question 16. Integrate \( e^{2x+3} \)
Answer: Let the integral be \( I = \int e^{2x+3} dx \). We should use a substitution here. Put \( 2x + 3 = t \). Differentiating both sides gives \( 2 \, dx = dt \), so \( dx = \frac{1}{2} dt \).
\( \implies I = \int e^{t} \left(\frac{1}{2} dt\right) \)
\( \implies I = \frac{1}{2} \int e^{t} dt \) The integral of \( e^{t} \) is \( e^{t} \).
\( \implies I = \frac{1}{2} e^{t} + C \) Substitute back \( t = 2x+3 \).
\( \implies I = \frac{1}{2} e^{2x+3} + C \).
In simple words: We let '2x + 3' be 't'. This means 'dx' becomes '1/2 dt'. We then integrate 'e to the power of t' and put '2x + 3' back in place of 't' to get the final solution.

Exam Tip: For exponential functions of the form \( e^{ax+b} \), the integral is \( \frac{1}{a}e^{ax+b} + C \). A substitution of \( t = ax+b \) helps to confirm this general formula.

Question 17. Integrate \( \frac{x}{e^{x^{2}}} \)
Answer: Let the integral be \( I = \int \frac{x}{e^{x^{2}}} dx \). We can rewrite the expression using a negative exponent.
\( \implies I = \int x e^{-x^{2}} dx \) Now, perform a substitution. Put \( x^{2} = t \). Differentiating both sides gives \( 2x \, dx = dt \), so \( x \, dx = \frac{1}{2} dt \).
\( \implies I = \int e^{-t} \left(\frac{1}{2} dt\right) \)
\( \implies I = \frac{1}{2} \int e^{-t} dt \) The integral of \( e^{-t} \) is \( -e^{-t} \).
\( \implies I = \frac{1}{2} (-e^{-t}) + C \)
\( \implies I = - \frac{1}{2} e^{-t} + C \) Substitute back \( t = x^{2} \).
\( \implies I = - \frac{1}{2} e^{-x^{2}} + C \) This can also be written as:
\( \implies I = - \frac{1}{2e^{x^{2}}} + C \).
In simple words: We rewrite the fraction by moving 'e to the power of x squared' to the numerator with a negative exponent. Then, we let 'x squared' be 't'. This changes 'x dx' to '1/2 dt'. We integrate 'e to the power of -t' and put 'x squared' back for 't' to get the solution.

Exam Tip: When dealing with \( e^{f(x)} \) and \( f'(x) \) is present in the integral, substitute \( t = f(x) \). Remember that \( e^{-t} \) integrates to \( -e^{-t} \).

Question 18. Integrate \( \frac{e^{\tan ^{-1} x}}{1+x^{2}} \)
Answer: Let the integral be \( I = \int \frac{e^{\tan^{-1}x}}{1+x^{2}} dx \). We should use a substitution here. Put \( \tan^{-1}x = t \). Differentiating both sides gives \( \frac{1}{1+x^{2}} dx = dt \).
\( \implies I = \int e^{t} dt \) The integral of \( e^{t} \) is \( e^{t} \).
\( \implies I = e^{t} + C \) Substitute back \( t = \tan^{-1}x \).
\( \implies I = e^{\tan^{-1}x} + C \).
In simple words: We let 'tan inverse x' be 't'. This changes '1 over (1 + x squared) dx' to 'dt'. We then integrate 'e to the power of t' and replace 't' with 'tan inverse x' for the final answer.

Exam Tip: Recognize the derivative of inverse trigonometric functions. The presence of \( \frac{1}{1+x^2} \) makes \( t = \tan^{-1}x \) an obvious and effective substitution.

Question 19. Integrate \( \frac{e^{2 x}-1}{e^{2x}+1} \)
Answer: Let the integral be \( I = \int \frac{e^{2x}-1}{e^{2x}+1} dx \). To simplify, divide the numerator and denominator by \( e^x \).
\( \implies I = \int \frac{(e^{2x}-1)/e^x}{(e^{2x}+1)/e^x} dx \)
\( \implies I = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx \) Now, perform a substitution. Put \( e^x + e^{-x} = t \). Differentiating both sides gives \( (e^x - e^{-x}) dx = dt \).
\( \implies I = \int \frac{dt}{t} \)
\( \implies I = \log |t| + C \) Substitute back \( t = e^x + e^{-x} \).
\( \implies I = \log |e^x + e^{-x}| + C \).
In simple words: We first divide the top and bottom of the fraction by 'e to the power of x'. Then, we make 'e to the power of x plus e to the power of -x' equal to 't'. This means the top part of the fraction becomes 'dt'. We then integrate '1 over t' and replace 't' with its original expression.

Exam Tip: For expressions involving \( e^{2x} \pm 1 \), a common trick is to divide the numerator and denominator by \( e^x \). This often transforms the integral into the form \( \int \frac{f'(x)}{f(x)} dx \).

Question 20. Integrate \( \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \)
Answer: Let the integral be \( I = \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} dx \). We should use a substitution here. Put \( e^{2x} + e^{-2x} = t \). Differentiating both sides gives \( (2e^{2x} - 2e^{-2x}) dx = dt \). Factor out 2: \( 2(e^{2x} - e^{-2x}) dx = dt \). This means \( (e^{2x} - e^{-2x}) dx = \frac{1}{2} dt \).
\( \implies I = \int \frac{1}{t} \left(\frac{1}{2} dt\right) \)
\( \implies I = \frac{1}{2} \int \frac{1}{t} dt \) The integral of \( \frac{1}{t} \) is \( \log |t| \).
\( \implies I = \frac{1}{2} \log |t| + C \) Substitute back \( t = e^{2x} + e^{-2x} \).
\( \implies I = \frac{1}{2} \log |e^{2x} + e^{-2x}| + C \).
In simple words: We let the bottom part of the fraction, 'e to the power of 2x plus e to the power of -2x', be 't'. This makes the top part of the fraction turn into '1/2 dt'. We then integrate '1 over t' and replace 't' with its original expression.

Exam Tip: When the numerator is the derivative (or a multiple of the derivative) of the denominator, use a substitution with \( t = \text{denominator} \). This simplifies the integral into a logarithmic form.

Question 21. Integrate \( \tan^{2}(2x - 3) \)
Answer: Let the integral be \( I = \int \tan^{2}(2x-3) dx \). We use the trigonometric identity \( \tan^{2}\theta = \sec^{2}\theta - 1 \).
\( \implies I = \int (\sec^{2}(2x-3) - 1) dx \)
\( \implies I = \int \sec^{2}(2x-3) dx - \int 1 \, dx \)
\( \implies I = \int \sec^{2}(2x-3) dx - x + C_1 \) Now, integrate the first term. Perform a substitution for \( (2x-3) \). Put \( 2x - 3 = t \). Differentiating both sides gives \( 2 \, dx = dt \), so \( dx = \frac{1}{2} dt \).
\( \implies \int \sec^{2}(2x-3) dx = \int \sec^{2}t \left(\frac{1}{2} dt\right) \)
\( \implies = \frac{1}{2} \int \sec^{2}t \, dt \) The integral of \( \sec^{2}t \) is \( \tan t \).
\( \implies = \frac{1}{2} \tan t + C_2 \) Substitute back \( t = 2x-3 \).
\( \implies = \frac{1}{2} \tan(2x-3) + C_2 \) Combine with the other term:
\( \implies I = \frac{1}{2} \tan(2x-3) - x + C \).
In simple words: We change 'tan squared' to 'sec squared minus 1'. Then we integrate each part separately. For the 'sec squared' part, we make '2x minus 3' equal to 't', integrate 'sec squared t', and put '2x minus 3' back.

Exam Tip: Remember to use trigonometric identities like \( \tan^2\theta = \sec^2\theta - 1 \) or \( \cot^2\theta = \csc^2\theta - 1 \) to transform squared tangent or cotangent functions into integrable forms.

Question 22. Integrate \( \sec^{2}(7 - 4x) \)
Answer: Let the integral be \( I = \int \sec^{2}(7-4x) dx \). We should use a substitution here. Put \( 7 - 4x = t \). Differentiating both sides gives \( -4 \, dx = dt \), so \( dx = - \frac{1}{4} dt \).
\( \implies I = \int \sec^{2}t \left(-\frac{1}{4} dt\right) \)
\( \implies I = - \frac{1}{4} \int \sec^{2}t \, dt \) The integral of \( \sec^{2}t \) is \( \tan t \).
\( \implies I = - \frac{1}{4} \tan t + C \) Substitute back \( t = 7-4x \).
\( \implies I = - \frac{1}{4} \tan(7-4x) + C \).
In simple words: We make '7 minus 4x' equal to 't'. This means 'dx' becomes '-1/4 dt'. We then integrate '-1/4 times sec squared t' and replace 't' with '7 minus 4x' to get the final answer.

Exam Tip: For standard trigonometric functions with a linear argument like \( \sec^2(ax+b) \), the integral will be \( \frac{1}{a}\tan(ax+b) + C \). A substitution helps to correctly manage the constant factor \( a \).

Question 23. Integrate \( \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \)
Answer: Let the integral be \( I = \int \frac{\sin^{-1}x}{\sqrt{1-x^{2}}} dx \). We should use a substitution here. Put \( \sin^{-1}x = t \). Differentiating both sides gives \( \frac{1}{\sqrt{1-x^{2}}} dx = dt \).
\( \implies I = \int t \, dt \) Using the power rule for integration:
\( \implies I = \frac{t^{1+1}}{1+1} + C \)
\( \implies I = \frac{t^{2}}{2} + C \) Substitute back \( t = \sin^{-1}x \).
\( \implies I = \frac{(\sin^{-1}x)^{2}}{2} + C \).
In simple words: We let 'sin inverse x' be 't'. This means '1 over sqrt(1 - x squared) dx' becomes 'dt'. We then integrate 't' and replace 't' with 'sin inverse x' to get the final answer.

Exam Tip: When an inverse trigonometric function appears alongside its derivative (or a multiple of it) in the integrand, a substitution using the inverse function as \( t \) is typically the correct approach.

Question 24. Integrate \( \frac{2\cos x-3\sin x}{6\cos x+4\sin x} \)
Answer: Let the integral be \( I = \int \frac{2\cos x-3\sin x}{6\cos x+4\sin x} dx \). First, factor out 2 from the denominator.
\( \implies I = \int \frac{2\cos x-3\sin x}{2(3\cos x+2\sin x)} dx \)
\( \implies I = \frac{1}{2} \int \frac{2\cos x-3\sin x}{3\cos x+2\sin x} dx \) Now, perform a substitution. Put \( 3\cos x + 2\sin x = t \). Differentiating both sides gives \( (-3\sin x + 2\cos x) dx = dt \). This is exactly the numerator.
\( \implies I = \frac{1}{2} \int \frac{dt}{t} \)
\( \implies I = \frac{1}{2} \log |t| + C \) Substitute back \( t = 3\cos x + 2\sin x \).
\( \implies I = \frac{1}{2} \log |3\cos x + 2\sin x| + C \).
In simple words: We first take out a '2' from the bottom part of the fraction. Then, we let the new denominator '3cos x + 2sin x' be 't'. This makes the top part of the fraction become 'dt'. We then integrate '1 over t' and replace 't' with its original expression.

Exam Tip: Always look for common factors in the denominator or numerator. If the numerator is the derivative (or a multiple of) the denominator, the integral will be a logarithmic function of the denominator.

Question 25. Integrate \( \frac{1}{\cos ^{2} x(1-\tan x)^{2}} \)
Answer: Let the integral be \( I = \int \frac{1}{\cos^{2}x(1-\tan x)^{2}} dx \). We know that \( \frac{1}{\cos^{2}x} = \sec^{2}x \).
\( \implies I = \int \frac{\sec^{2}x}{(1-\tan x)^{2}} dx \) Now, perform a substitution. Put \( 1 - \tan x = t \). Differentiating both sides gives \( -\sec^{2}x \, dx = dt \), so \( \sec^{2}x \, dx = -dt \).
\( \implies I = \int \frac{-dt}{t^{2}} \)
\( \implies I = - \int t^{-2} dt \) Using the power rule for integration:
\( \implies I = - \left( \frac{t^{-2+1}}{-2+1} \right) + C \)
\( \implies I = - \left( \frac{t^{-1}}{-1} \right) + C \)
\( \implies I = t^{-1} + C \)
\( \implies I = \frac{1}{t} + C \) Substitute back \( t = 1-\tan x \).
\( \implies I = \frac{1}{1-\tan x} + C \).
In simple words: We first change '1 over cos squared x' to 'sec squared x'. Then, we let '1 minus tan x' be 't'. This makes 'sec squared x dx' become '-dt'. We then integrate '-1 over t squared' and replace 't' with '1 minus tan x' to find the answer.

Exam Tip: When you see \( \sec^2 x \) and \( \tan x \) in an integral, consider substituting \( t = \tan x \) or \( t = (1-\tan x) \) because \( \sec^2 x \) is the derivative of \( \tan x \).

Question 26. Integrate \( \cot x \log \sin x \)
Answer: Let the integral be \( I = \int \cot x \log \sin x \, dx \). We should use a substitution here. Put \( \log \sin x = t \). Differentiating both sides gives \( \frac{1}{\sin x} (\cos x) dx = dt \).
\( \implies \cot x \, dx = dt \).
\( \implies I = \int t \, dt \) Using the power rule for integration:
\( \implies I = \frac{t^{1+1}}{1+1} + C \)
\( \implies I = \frac{t^{2}}{2} + C \) Substitute back \( t = \log \sin x \).
\( \implies I = \frac{(\log \sin x)^{2}}{2} + C \).
In simple words: We let 'log sin x' be 't'. This changes 'cot x dx' to 'dt'. We then integrate 't' and replace 't' with 'log sin x' to get the final solution.

Exam Tip: If you spot a function (like \( \log \sin x \)) and its derivative (like \( \cot x \)) within the integral, a simple substitution is usually the most effective method.

Question 27. Integrate \( \sqrt{\sin 2x} \cos 2x \)
Answer: Let the integral be \( I = \int \sqrt{\sin 2x} \cos 2x \, dx \). We should use a substitution here. Put \( \sin 2x = t \). Differentiating both sides gives \( (\cos 2x)(2) dx = dt \), so \( \cos 2x \, dx = \frac{1}{2} dt \).
\( \implies I = \int \sqrt{t} \left(\frac{1}{2} dt\right) \)
\( \implies I = \frac{1}{2} \int t^{1/2} dt \) Using the power rule for integration:
\( \implies I = \frac{1}{2} \left( \frac{t^{1/2+1}}{1/2+1} \right) + C \)
\( \implies I = \frac{1}{2} \left( \frac{t^{3/2}}{3/2} \right) + C \)
\( \implies I = \frac{1}{2} \left( \frac{2}{3} t^{3/2} \right) + C \)
\( \implies I = \frac{1}{3} t^{3/2} + C \) Substitute back \( t = \sin 2x \).
\( \implies I = \frac{1}{3} (\sin 2x)^{3/2} + C \).
In simple words: We let 'sin 2x' be 't'. This means 'cos 2x dx' becomes '1/2 dt'. We then integrate '1/2 times sqrt t' and replace 't' with 'sin 2x' to get the final answer.

Exam Tip: When integrating functions involving \( \sqrt{f(x)} \cdot f'(x) \), a substitution of \( t=f(x) \) is generally the quickest method. Remember to handle the constant factors correctly.

Question 28. Integrate \( \frac{\cos x}{\sqrt{1+\sin x}} \)
Answer: Let the integral be \( I = \int \frac{\cos x}{\sqrt{1+\sin x}} dx \). We should use a substitution here. Put \( 1 + \sin x = t \). Differentiating both sides gives \( \cos x \, dx = dt \).
\( \implies I = \int \frac{1}{\sqrt{t}} dt \)
\( \implies I = \int t^{-1/2} dt \) Using the power rule for integration:
\( \implies I = \frac{t^{-1/2+1}}{-1/2+1} + C \)
\( \implies I = \frac{t^{1/2}}{1/2} + C \)
\( \implies I = 2t^{1/2} + C \)
\( \implies I = 2\sqrt{t} + C \) Substitute back \( t = 1+\sin x \).
\( \implies I = 2\sqrt{1+\sin x} + C \).
In simple words: We let '1 + sin x' be 't'. This changes 'cos x dx' to 'dt'. We then integrate '1 over sqrt t' and replace 't' with '1 + sin x' to get the final answer.

Exam Tip: For integrals of the form \( \frac{f'(x)}{\sqrt{f(x)}} \), a substitution \( t=f(x) \) simplifies the integral to \( \int t^{-1/2} dt \), which is easily solved using the power rule.

Question 29. Integrate \( \cot x \log \sin x \)
Answer: Let the integral be \( I = \int \cot x \log \sin x \, dx \). We should use a substitution here. Put \( \log \sin x = t \). Differentiating both sides gives \( \frac{1}{\sin x} (\cos x) dx = dt \).
\( \implies \cot x \, dx = dt \).
\( \implies I = \int t \, dt \) Using the power rule for integration:
\( \implies I = \frac{t^{1+1}}{1+1} + C \)
\( \implies I = \frac{t^{2}}{2} + C \) Substitute back \( t = \log \sin x \).
\( \implies I = \frac{(\log \sin x)^{2}}{2} + C \).
In simple words: We let 'log sin x' be 't'. This changes 'cot x dx' to 'dt'. We then integrate 't' and replace 't' with 'log sin x' to get the final solution.

Exam Tip: If you spot a function (like \( \log \sin x \)) and its derivative (like \( \cot x \)) within the integral, a simple substitution is usually the most effective method.

Question 30. Integrate \( \frac{\sin x}{1+\cos x} \)
Answer: Let the integral be \( I = \int \frac{\sin x}{1+\cos x} dx \). We should use a substitution here. Put \( 1 + \cos x = t \). Differentiating both sides gives \( -\sin x \, dx = dt \), so \( \sin x \, dx = -dt \).
\( \implies I = \int \frac{-dt}{t} \)
\( \implies I = - \int \frac{1}{t} dt \) The integral of \( \frac{1}{t} \) is \( \log |t| \).
\( \implies I = - \log |t| + C \) Substitute back \( t = 1+\cos x \).
\( \implies I = - \log |1+\cos x| + C \).
In simple words: We make '1 + cos x' equal to 't'. This changes 'sin x dx' to '-dt'. We then integrate '-1 over t' and replace 't' with '1 + cos x' to get the final answer.

Exam Tip: For rational functions where the numerator is the negative of the derivative of the denominator, a substitution \( t = \text{denominator} \) leads to a logarithmic integral.

Question 31. Integrate \( \frac{\sin x}{(1+\cos x)^{2}} \)
Answer: Let the integral be \( I = \int \frac{\sin x}{(1+\cos x)^{2}} dx \). We should use a substitution here. Put \( 1 + \cos x = t \). Differentiating both sides gives \( -\sin x \, dx = dt \), so \( \sin x \, dx = -dt \).
\( \implies I = \int \frac{-dt}{t^{2}} \)
\( \implies I = - \int t^{-2} dt \) Using the power rule for integration:
\( \implies I = - \left( \frac{t^{-2+1}}{-2+1} \right) + C \)
\( \implies I = - \left( \frac{t^{-1}}{-1} \right) + C \)
\( \implies I = t^{-1} + C \)
\( \implies I = \frac{1}{t} + C \) Substitute back \( t = 1+\cos x \).
\( \implies I = \frac{1}{1+\cos x} + C \).
In simple words: We let '1 + cos x' be 't'. This changes 'sin x dx' to '-dt'. We then integrate '-1 over t squared' and replace 't' with '1 + cos x' to find the answer.

Exam Tip: Look for patterns involving a function and its derivative. Here, \( \sin x \) is related to the derivative of \( (1+\cos x) \), making substitution a suitable method.

Question 32. Integrate \( \frac{1}{1+\cot x} \)
Answer: Let the integral be \( I = \int \frac{1}{1+\cot x} dx \). First, rewrite \( \cot x \) as \( \frac{\cos x}{\sin x} \).
\( \implies I = \int \frac{1}{1+\frac{\cos x}{\sin x}} dx \) Combine the terms in the denominator.
\( \implies I = \int \frac{1}{\frac{\sin x+\cos x}{\sin x}} dx \)
\( \implies I = \int \frac{\sin x}{\sin x+\cos x} dx \) Now, we use a common trick. Multiply and divide by 2, and add/subtract \( \cos x \) in the numerator.
\( \implies I = \frac{1}{2} \int \frac{2\sin x}{\sin x+\cos x} dx \)
\( \implies I = \frac{1}{2} \int \frac{(\sin x+\cos x) + (\sin x-\cos x)}{\sin x+\cos x} dx \) Split the integral into two parts.
\( \implies I = \frac{1}{2} \int \left( \frac{\sin x+\cos x}{\sin x+\cos x} + \frac{\sin x-\cos x}{\sin x+\cos x} \right) dx \)
\( \implies I = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} dx \)
\( \implies I = \frac{1}{2} x + C_1 + \frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} dx \) For the second integral, let's substitute. Let \( \sin x + \cos x = t \). Differentiating both sides gives \( (\cos x - \sin x) dx = dt \), so \( (\sin x - \cos x) dx = -dt \).
\( \implies \frac{1}{2} \int \frac{-dt}{t} = - \frac{1}{2} \int \frac{1}{t} dt \)
\( \implies = - \frac{1}{2} \log |t| + C_2 \)
\( \implies = - \frac{1}{2} \log |\sin x + \cos x| + C_2 \) Combine both parts.
\( \implies I = \frac{x}{2} - \frac{1}{2} \log |\sin x + \cos x| + C \).
In simple words: First, we change 'cot x' to 'cos x over sin x'. Then, we combine the bottom part of the fraction. We split the integral into two simpler parts, one for '1' and one for a fraction where the top is almost the derivative of the bottom. We use substitution for the second part and then add everything together.

Exam Tip: For integrals of the form \( \int \frac{1}{1 \pm \tan x} dx \) or \( \int \frac{1}{1 \pm \cot x} dx \), convert tangent/cotangent to sine/cosine. Then, use the trick of writing the numerator as a combination of the denominator and its derivative.

Question 33. Integrate \( \frac{1}{1-\tan x} \)
Answer: Let the integral be \( I = \int \frac{1}{1-\tan x} dx \). First, rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \).
\( \implies I = \int \frac{1}{1-\frac{\sin x}{\cos x}} dx \) Combine the terms in the denominator.
\( \implies I = \int \frac{1}{\frac{\cos x-\sin x}{\cos x}} dx \)
\( \implies I = \int \frac{\cos x}{\cos x-\sin x} dx \) Now, use the trick: multiply and divide by 2, and add/subtract \( \sin x \) in the numerator.
\( \implies I = \frac{1}{2} \int \frac{2\cos x}{\cos x-\sin x} dx \)
\( \implies I = \frac{1}{2} \int \frac{(\cos x-\sin x) + (\cos x+\sin x)}{\cos x-\sin x} dx \) Split the integral into two parts.
\( \implies I = \frac{1}{2} \int \left( \frac{\cos x-\sin x}{\cos x-\sin x} + \frac{\cos x+\sin x}{\cos x-\sin x} \right) dx \)
\( \implies I = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} dx \)
\( \implies I = \frac{1}{2} x + C_1 + \frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} dx \) For the second integral, let's substitute. Let \( \cos x - \sin x = t \). Differentiating both sides gives \( (-\sin x - \cos x) dx = dt \), so \( (\cos x + \sin x) dx = -dt \).
\( \implies \frac{1}{2} \int \frac{-dt}{t} = - \frac{1}{2} \int \frac{1}{t} dt \)
\( \implies = - \frac{1}{2} \log |t| + C_2 \)
\( \implies = - \frac{1}{2} \log |\cos x - \sin x| + C_2 \) Combine both parts.
\( \implies I = \frac{x}{2} - \frac{1}{2} \log |\cos x - \sin x| + C \).
In simple words: We first change 'tan x' to 'sin x over cos x' and combine the bottom part. We then split the integral into two parts, one for '1' and one for a fraction where the top is related to the derivative of the bottom. We use substitution for the second part and then add the results.

Exam Tip: Similar to \( 1+\cot x \), this integral requires converting \( \tan x \) to \( \sin x/\cos x \) and then manipulating the numerator to create terms that integrate easily, including a logarithmic term.

Question 34. Integrate \( \frac{\sqrt{\tan x}}{\sin x \cos x} \)
Answer: Let the integral be \( I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx \). To make a \( \tan x \) term, divide the denominator by \( \cos x \) and multiply by \( \cos x \).
\( \implies I = \int \frac{\sqrt{\tan x}}{(\frac{\sin x}{\cos x}) \cos^{2}x} dx \)
\( \implies I = \int \frac{\sqrt{\tan x}}{\tan x \cos^{2}x} dx \) We know \( \frac{1}{\cos^{2}x} = \sec^{2}x \).
\( \implies I = \int \frac{\sqrt{\tan x}}{\tan x} \sec^{2}x \, dx \)
\( \implies I = \int (\tan x)^{-1/2} \sec^{2}x \, dx \) Now, perform a substitution. Put \( \tan x = t \). Differentiating both sides gives \( \sec^{2}x \, dx = dt \).
\( \implies I = \int t^{-1/2} dt \) Using the power rule for integration:
\( \implies I = \frac{t^{-1/2+1}}{-1/2+1} + C \)
\( \implies I = \frac{t^{1/2}}{1/2} + C \)
\( \implies I = 2t^{1/2} + C \)
\( \implies I = 2\sqrt{t} + C \) Substitute back \( t = \tan x \).
\( \implies I = 2\sqrt{\tan x} + C \).
In simple words: We first change the fraction to involve 'tan x' and 'sec squared x'. We do this by dividing and multiplying by 'cos x'. Then, we let 'tan x' be 't'. This changes 'sec squared x dx' to 'dt'. We then integrate 't to the power of -1/2' and replace 't' with 'tan x' to find the answer.

Exam Tip: When \( \tan x \) is under a root in the numerator and \( \sin x \cos x \) is in the denominator, multiply the denominator by \( \cos x/\cos x \) to create \( \tan x \) and \( \sec^2 x \), which are perfect for substitution.

Question 35. Integrate \( \frac{(1+\log x)^{2}}{x} \)
Answer: Let the integral be \( I = \int \frac{(1+\log x)^{2}}{x} dx \). We should use a substitution here. Put \( 1 + \log x = t \). Differentiating both sides gives \( \frac{1}{x} dx = dt \).
\( \implies I = \int t^{2} dt \) Using the power rule for integration:
\( \implies I = \frac{t^{2+1}}{2+1} + C \)
\( \implies I = \frac{t^{3}}{3} + C \) Substitute back \( t = 1+\log x \).
\( \implies I = \frac{(1+\log x)^{3}}{3} + C \).
In simple words: We let '1 + log x' be 't'. This means '1/x dx' becomes 'dt'. We then integrate 't squared' and replace 't' with '1 + log x' to get the final answer.

Exam Tip: The presence of \( \log x \) and \( \frac{1}{x} \) together is a strong indicator for a substitution where \( t \) is related to \( \log x \).

Question 36. Integrate \( \frac{(x+1)(x+\log x)^{2}}{x} \)
Answer: Let the integral be \( I = \int \frac{(x+1)(x+\log x)^{2}}{x} dx \). We can rearrange the expression to make a substitution easier.
\( \implies I = \int (x+\log x)^{2} \frac{(x+1)}{x} dx \)
\( \implies I = \int (x+\log x)^{2} \left(1+\frac{1}{x}\right) dx \) Now, perform a substitution. Put \( x + \log x = t \). Differentiating both sides gives \( \left(1+\frac{1}{x}\right) dx = dt \).
\( \implies I = \int t^{2} dt \) Using the power rule for integration:
\( \implies I = \frac{t^{2+1}}{2+1} + C \)
\( \implies I = \frac{t^{3}}{3} + C \) Substitute back \( t = x+\log x \).
\( \implies I = \frac{(x+\log x)^{3}}{3} + C \).
In simple words: We first rewrite the fraction to put '(1 + 1/x)' next to the 'x + log x' term. Then, we let 'x + log x' be 't'. This makes '(1 + 1/x) dx' become 'dt'. We then integrate 't squared' and replace 't' with 'x + log x' to get the final answer.

Exam Tip: When you have a complex function raised to a power and another term multiplying it, check if that multiplier is the derivative of the complex function. Here, \( 1+\frac{1}{x} \) is the derivative of \( x+\log x \).

Question 37. Integrate \( \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}} \)
Answer: Let the integral be \( I = \int \frac{x^{3} \sin \left(\tan^{-1}x^{4}\right)}{1+x^{8}} dx \). We should use a substitution here. Put \( \tan^{-1}x^{4} = t \). Differentiating both sides gives \( \frac{1}{1+(x^{4})^{2}} \cdot (4x^{3}) dx = dt \).
\( \implies \frac{4x^{3}}{1+x^{8}} dx = dt \)
\( \implies \frac{x^{3}}{1+x^{8}} dx = \frac{1}{4} dt \).
\( \implies I = \int \sin(t) \left(\frac{1}{4} dt\right) \)
\( \implies I = \frac{1}{4} \int \sin t \, dt \) The integral of \( \sin t \) is \( -\cos t \).
\( \implies I = \frac{1}{4} (-\cos t) + C \)
\( \implies I = - \frac{1}{4} \cos t + C \) Substitute back \( t = \tan^{-1}x^{4} \).
\( \implies I = - \frac{1}{4} \cos(\tan^{-1}x^{4}) + C \).
In simple words: We make 'tan inverse x to the power of 4' equal to 't'. This changes 'x cubed over (1 + x to the power of 8) dx' to '1/4 dt'. We then integrate 'sin t' and replace 't' with 'tan inverse x to the power of 4' to find the answer.

Exam Tip: Look for composite functions where the derivative of the inner function (or a part of it) is also present. The derivative of \( \tan^{-1}(f(x)) \) involves \( \frac{f'(x)}{1+(f(x))^2} \), which is evident here.

Question 38. \( \int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}dx \) equals
(a) \( 10x^9 + 10^x \log_e 10 + C \)
(b) \( 10^x + x^{10} + C \)
(c) \( (10^x – x^{10})^{-1} + C \)
(d) \( \log(10^x + x^{10}) + C \)
Answer: (d) \( \log(10^x + x^{10}) + C \)
In simple words: The integral is in the form of f'(x) over f(x), which means the answer will be the natural logarithm of the bottom part.

Exam Tip: Always examine the relationship between the numerator and denominator in a rational integral. If the numerator is the derivative of the denominator, the integral is simply the natural logarithm of the absolute value of the denominator.

Question 39. \( \int \frac{d x}{\sin ^{2} x \cos ^{2} x} \) equals to
(A) \( \tan x + \cot x + C \)
(B) \( \tan x - \cot x + C \)
(C) \( \tan x \cot x + C \)
(D) \( \tan x \cot x + C \)
Answer: (B) \( \tan x - \cot x + C \)
In simple words: We replace '1' in the numerator with 'sin squared x plus cos squared x'. Then we separate the fraction into two parts, which simplify to 'sec squared x' and 'cosec squared x'. Integrating these gives 'tan x minus cot x'.

Exam Tip: When the denominator involves products of \( \sin^2 x \) and \( \cos^2 x \), a common strategy is to replace the numerator '1' with the identity \( \sin^2 x + \cos^2 x \) and then split the fraction to simplify and integrate.