GSEB Class 12 Maths Solutions Chapter 7 સંકલન Exercise 7.6

Question 1. \( x \sin x \)
Answer:
\( I = \int x \sin x \, dx \)
Here, we take \( u = x \) and \( v = \sin x \). Then, applying the rule of integration by parts, we get:
\( I = x \int \sin x \, dx - \int \left( \frac{d}{dx}(x) \int \sin x \, dx \right) \, dx \)
\( = -x \cos x - \int (- \cos x) \, dx \)
\( = -x \cos x + \sin x + C \)
In simple words: To integrate this, we use the "integration by parts" method. We choose \( x \) as \( u \) and \( \sin x \) as \( v \). After applying the formula, the final answer comes out as \( -x \cos x + \sin x + C \).

Exam Tip: Remember the LIATE rule for choosing 'u' and 'v' in integration by parts. Here, algebraic \( x \) comes before trigonometric \( \sin x \).

Question 2. \( x \sin 3x \)
Answer:
\( I = \int x \sin 3x \, dx \)
Here, we choose \( u = x \) and \( v = \sin 3x \). Applying the integration by parts formula, we get:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = x \int \sin 3x \, dx - \int \left( \frac{d}{dx}(x) \int \sin 3x \, dx \right) \, dx \)
\( = x \left( \frac{-\cos 3x}{3} \right) - \int (1) \left( \frac{-\cos 3x}{3} \right) \, dx \)
\( = \frac{-x \cos 3x}{3} + \frac{1}{3} \int \cos 3x \, dx \)
\( = \frac{-x \cos 3x}{3} + \frac{1}{3} \left( \frac{\sin 3x}{3} \right) + C \)
\( = \frac{-x \cos 3x}{3} + \frac{\sin 3x}{9} + C \)
In simple words: This problem also uses integration by parts. We set \( u = x \) and \( v = \sin 3x \). We then follow the formula to find the integral, which involves integrating \( \sin 3x \) and also differentiating \( x \).

Exam Tip: Always make sure to divide by the coefficient of x when integrating trigonometric functions like \( \sin(ax+b) \) or \( \cos(ax+b) \).

Question 3. \( x^2 e^x \)
Answer:
\( I = \int x^2 e^x \, dx \)
We take \( u = x^2 \) and \( v = e^x \). Using the rule for integration by parts, we find:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = x^2 \int e^x \, dx - \int \left( \frac{d}{dx}(x^2) \int e^x \, dx \right) \, dx \)
\( = x^2 e^x - \int 2x e^x \, dx \)
\( = x^2 e^x - 2 \int x e^x \, dx \)
Now, we apply integration by parts again for \( \int x e^x \, dx \), using \( u = x \) and \( v = e^x \):
\( = x^2 e^x - 2 \left[ x \int e^x \, dx - \int \left( \frac{d}{dx}(x) \int e^x \, dx \right) \, dx \right] \)
\( = x^2 e^x - 2 [x e^x - \int 1 \cdot e^x \, dx] \)
\( = x^2 e^x - 2 [x e^x - e^x] + C \)
\( = x^2 e^x - 2x e^x + 2 e^x + C \)
\( = e^x (x^2 - 2x + 2) + C \)
In simple words: This integral needs the integration by parts rule to be applied twice. First, we integrate \( x^2 e^x \), which then gives us a new integral \( \int x e^x \, dx \). We then apply the same method to solve this new integral to get the final answer.

Exam Tip: When \( x \) is raised to a power greater than 1, you often need to apply integration by parts multiple times until the \( x \) term becomes 1.

Question 4. \( x \log x \)
Answer:
\( I = \int x \log x \, dx \)
We choose \( u = \log x \) and \( v = x \). Using the formula for integration by parts, we get:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = \log x \int x \, dx - \int \left( \frac{d}{dx}(\log x) \int x \, dx \right) \, dx \)
\( = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2}{2} \log x - \int \frac{x}{2} \, dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx \)
\( = \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} + C \)
\( = \frac{x^2}{2} \log x - \frac{x^2}{4} + C \)
In simple words: To solve this integral, we use the integration by parts rule. We select \( \log x \) as \( u \) and \( x \) as \( v \). After applying the formula, the solution ends up being \( \frac{x^2}{2} \log x - \frac{x^2}{4} + C \).

Exam Tip: When integrating terms with \( \log x \), always choose \( \log x \) as 'u' as it simplifies upon differentiation.

Question 5. \( x \log 2x \)
Answer:
\( I = \int x \log 2x \, dx \)
We take \( u = \log 2x \) and \( v = x \). Using the rule for integration by parts, we find:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = \log 2x \int x \, dx - \int \left( \frac{d}{dx}(\log 2x) \int x \, dx \right) \, dx \)
\( = \log 2x \cdot \frac{x^2}{2} - \int \frac{1}{2x} \cdot 2 \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2}{2} \log 2x - \int \frac{x}{2} \, dx \)
\( = \frac{x^2}{2} \log 2x - \frac{1}{2} \int x \, dx \)
\( = \frac{x^2}{2} \log 2x - \frac{1}{2} \cdot \frac{x^2}{2} + C \)
\( = \frac{x^2}{2} \log 2x - \frac{x^2}{4} + C \)
In simple words: For this integral, we apply integration by parts. We choose \( \log 2x \) as \( u \) and \( x \) as \( v \). When we follow the formula, the result we get is \( \frac{x^2}{2} \log 2x - \frac{x^2}{4} + C \).

Exam Tip: Be careful when differentiating \( \log(ax) \), applying the chain rule correctly as \( \frac{1}{ax} \cdot a = \frac{1}{x} \).

Question 6. \( x^2 \log x \)
Answer:
\( I = \int x^2 \log x \, dx \)
We take \( u = \log x \) and \( v = x^2 \). Using the integration by parts rule, we find:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = \log x \int x^2 \, dx - \int \left( \frac{d}{dx}(\log x) \int x^2 \, dx \right) \, dx \)
\( = \log x \cdot \frac{x^3}{3} - \int \frac{1}{x} \cdot \frac{x^3}{3} \, dx \)
\( = \frac{x^3}{3} \log x - \int \frac{x^2}{3} \, dx \)
\( = \frac{x^3}{3} \log x - \frac{1}{3} \int x^2 \, dx \)
\( = \frac{x^3}{3} \log x - \frac{1}{3} \cdot \frac{x^3}{3} + C \)
\( = \frac{x^3}{3} \log x - \frac{x^3}{9} + C \)
In simple words: To integrate this, we use the integration by parts method. We select \( \log x \) as \( u \) and \( x^2 \) as \( v \). Following the formula, the answer we achieve is \( \frac{x^3}{3} \log x - \frac{x^3}{9} + C \).

Exam Tip: Always prioritize logarithmic functions for 'u' in integration by parts (LIATE rule) as their derivatives simplify.

Question 7. \( x \sin^{-1} x \)
Answer:
\( I = \int x \sin^{-1} x \, dx \)
We choose \( u = \sin^{-1} x \) and \( v = x \). Applying the integration by parts formula:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = \sin^{-1} x \int x \, dx - \int \left( \frac{d}{dx}(\sin^{-1} x) \int x \, dx \right) \, dx \)
\( = \sin^{-1} x \cdot \frac{x^2}{2} - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx \)
Let \( I_1 = \int \frac{x^2}{\sqrt{1-x^2}} \, dx \). So, \( I = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} I_1 \) ....(i)
To solve \( I_1 \), let \( x = \sin \theta \), which means \( dx = \cos \theta \, d\theta \). Then \( \theta = \sin^{-1} x \).
\( I_1 = \int \frac{\sin^2 \theta}{\sqrt{1-\sin^2 \theta}} \cos \theta \, d\theta \)
\( = \int \frac{\sin^2 \theta}{\sqrt{\cos^2 \theta}} \cos \theta \, d\theta \)
\( = \int \frac{\sin^2 \theta}{\cos \theta} \cos \theta \, d\theta \)
\( = \int \sin^2 \theta \, d\theta \)
\( = \int \frac{1 - \cos 2\theta}{2} \, d\theta \)
\( = \frac{1}{2} \int (1 - \cos 2\theta) \, d\theta \)
\( = \frac{1}{2} \left[ \theta - \frac{\sin 2\theta}{2} \right] + C_1 \)
\( = \frac{1}{2} \left[ \theta - \frac{2 \sin \theta \cos \theta}{2} \right] + C_1 \)
\( = \frac{1}{2} [\theta - \sin \theta \cos \theta] + C_1 \)
\( = \frac{1}{2} [\theta - \sin \theta \sqrt{1-\sin^2 \theta}] + C_1 \)
Substitute back \( x = \sin \theta \) and \( \theta = \sin^{-1} x \):
\( I_1 = \frac{1}{2} [\sin^{-1} x - x \sqrt{1-x^2}] + C_1 \)
Now, substitute the value of \( I_1 \) into equation (i):
\( I = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \left( \frac{1}{2} [\sin^{-1} x - x \sqrt{1-x^2}] \right) + C \)
\( = \frac{x^2}{2} \sin^{-1} x - \frac{1}{4} \sin^{-1} x + \frac{x}{4} \sqrt{1-x^2} + C \)
\( = \sin^{-1} x \left( \frac{x^2}{2} - \frac{1}{4} \right) + \frac{x}{4} \sqrt{1-x^2} + C \)
\( = \sin^{-1} x \left( \frac{2x^2 - 1}{4} \right) + \frac{x}{4} \sqrt{1-x^2} + C \)
In simple words: This problem needs integration by parts. We first set \( u = \sin^{-1} x \) and \( v = x \). This leads us to another integral, \( I_1 \), which we solve by substituting \( x = \sin \theta \). After finding \( I_1 \), we put it back into the main equation to get the full answer.

Exam Tip: Integrals involving inverse trigonometric functions often require substitution (e.g., \( x = \sin \theta \) or \( x = \tan \theta \)) after applying integration by parts once.

Question 8. \( x \tan^{-1} x \)
Answer:
\( I = \int x \tan^{-1} x \, dx \)
We take \( u = \tan^{-1} x \) and \( v = x \). Using the rule of integration by parts, we get:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = \tan^{-1} x \int x \, dx - \int \left( \frac{d}{dx}(\tan^{-1} x) \int x \, dx \right) \, dx \)
\( = \tan^{-1} x \cdot \frac{x^2}{2} - \int \frac{1}{1+x^2} \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2+1-1}{1+x^2} \, dx \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) \, dx \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \left[ \int 1 \, dx - \int \frac{1}{1+x^2} \, dx \right] \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} [x - \tan^{-1} x] + C \)
\( = \frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + \frac{1}{2} \tan^{-1} x + C \)
In simple words: This problem involves integration by parts. We choose \( \tan^{-1} x \) as \( u \) and \( x \) as \( v \). After applying the formula, we simplify the resulting integral by adding and subtracting 1 in the numerator. This helps us to integrate it directly.

Exam Tip: For integrals of the form \( \int \frac{P(x)}{Q(x)} \, dx \) where degree of P(x) is greater than or equal to degree of Q(x), use polynomial long division or add/subtract terms to simplify the integrand.

Question 9. \( x \cos^{-1} x \)
Answer:
\( I = \int x \cos^{-1} x \, dx \)
We take \( u = \cos^{-1} x \) and \( v = x \). Using the formula for integration by parts, we get:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = \cos^{-1} x \int x \, dx - \int \left( \frac{d}{dx}(\cos^{-1} x) \int x \, dx \right) \, dx \)
\( = \cos^{-1} x \cdot \frac{x^2}{2} - \int \frac{-1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx \)
Let \( I_1 = \int \frac{x^2}{\sqrt{1-x^2}} \, dx \). So, \( I = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} I_1 \) ....(i)
To solve \( I_1 \), let \( x = \cos \theta \), which means \( dx = -\sin \theta \, d\theta \). Then \( \theta = \cos^{-1} x \).
\( I_1 = \int \frac{\cos^2 \theta}{\sqrt{1-\cos^2 \theta}} (-\sin \theta) \, d\theta \)
\( = \int \frac{\cos^2 \theta}{\sqrt{\sin^2 \theta}} (-\sin \theta) \, d\theta \)
\( = \int \frac{\cos^2 \theta}{\sin \theta} (-\sin \theta) \, d\theta \)
\( = \int -\cos^2 \theta \, d\theta \)
\( = - \int \frac{1 + \cos 2\theta}{2} \, d\theta \)
\( = -\frac{1}{2} \int (1 + \cos 2\theta) \, d\theta \)
\( = -\frac{1}{2} \left[ \theta + \frac{\sin 2\theta}{2} \right] + C_1 \)
\( = -\frac{1}{2} \left[ \theta + \frac{2 \sin \theta \cos \theta}{2} \right] + C_1 \)
\( = -\frac{1}{2} [\theta + \sin \theta \cos \theta] + C_1 \)
\( = -\frac{1}{2} [\theta + \sqrt{1-\cos^2 \theta} \cdot \cos \theta] + C_1 \)
Substitute back \( x = \cos \theta \) and \( \theta = \cos^{-1} x \):
\( I_1 = -\frac{1}{2} [\cos^{-1} x + x \sqrt{1-x^2}] + C_1 \)
Now, substitute the value of \( I_1 \) into equation (i):
\( I = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \left( -\frac{1}{2} [\cos^{-1} x + x \sqrt{1-x^2}] \right) + C \)
\( = \frac{x^2}{2} \cos^{-1} x - \frac{1}{4} \cos^{-1} x - \frac{x}{4} \sqrt{1-x^2} + C \)
\( = \cos^{-1} x \left( \frac{x^2}{2} - \frac{1}{4} \right) - \frac{x}{4} \sqrt{1-x^2} + C \)
\( = \cos^{-1} x \left( \frac{2x^2 - 1}{4} \right) - \frac{x}{4} \sqrt{1-x^2} + C \)
In simple words: This integral uses integration by parts, with \( u = \cos^{-1} x \) and \( v = x \). We solve the resulting integral \( I_1 \) by substituting \( x = \cos \theta \). After that, we substitute \( I_1 \) back into the initial equation to get the complete solution.

Exam Tip: Be careful with the negative sign when differentiating \( \cos^{-1} x \), which is \( \frac{-1}{\sqrt{1-x^2}} \).

Question 11. \( \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \)
Answer:
\( I = \int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} \, dx \)
Let \( \cos^{-1} x = t \). Then, differentiating both sides, \( \frac{-1}{\sqrt{1-x^2}} \, dx = dt \). This also means \( x = \cos t \).
So, \( I = \int (\cos t) \cdot t \cdot (-dt) \)
\( = - \int t \cos t \, dt \)
Now, we apply integration by parts to \( \int t \cos t \, dt \). We take \( u = t \) and \( v = \cos t \):
\( \int uv \, dt = u \int v \, dt - \int \left( \frac{d}{dt}(u) \int v \, dt \right) \, dt \)
\( = - \left[ t \int \cos t \, dt - \int \left( \frac{d}{dt}(t) \int \cos t \, dt \right) \, dt \right] \)
\( = - [t \sin t - \int (1) \sin t \, dt] \)
\( = - [t \sin t - (-\cos t)] + C \)
\( = - [t \sin t + \cos t] + C \)
\( = - t \sin t - \cos t + C \)
\( = - t \sqrt{1-\cos^2 t} - \cos t + C \)
Substitute back \( t = \cos^{-1} x \) and \( x = \cos t \):
\( = - \cos^{-1} x \cdot \sqrt{1-x^2} - x + C \)
\( = - [\cos^{-1} x \sqrt{1-x^2} + x] + C \)
In simple words: To integrate this expression, we first use a substitution by letting \( t = \cos^{-1} x \). This transforms the integral into a simpler form involving \( t \cos t \). Then, we solve this new integral using the integration by parts method. Finally, we replace \( t \) back with \( \cos^{-1} x \) to get the solution.

Exam Tip: For integrals of the form \( \int f(x) \cdot g'(x) \, dx \) where \( g'(x) \) is part of the differential, substitution is often more efficient than direct integration by parts.

Question 12. \( x \sec^2 x \)
Answer:
\( I = \int x \sec^2 x \, dx \)
We take \( u = x \) and \( v = \sec^2 x \). Applying the rules of integration by parts, we find:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = x \int \sec^2 x \, dx - \int \left( \frac{d}{dx}(x) \int \sec^2 x \, dx \right) \, dx \)
\( = x \tan x - \int (1) \tan x \, dx \)
\( = x \tan x - \int \tan x \, dx \)
\( = x \tan x - (-\log |\cos x|) + C \)
\( = x \tan x + \log |\cos x| + C \)
(Since \( \int \frac{f'(x)}{f(x)} \, dx = \log |f(x)| + C \))
In simple words: We solve this integral using the integration by parts technique. We choose \( x \) as \( u \) and \( \sec^2 x \) as \( v \). After performing the required steps, the integral of \( \tan x \) gives us \( -\log |\cos x| \), leading to the final result.

Exam Tip: Remember the integral of \( \tan x \) is \( -\log |\cos x| \) or \( \log |\sec x| \). Use the LIATE rule to correctly choose 'u' and 'v'.

Question 13. \( \tan^{-1} x \)
Answer:
\( I = \int \tan^{-1} x \, dx \)
We can write this as \( I = \int \tan^{-1} x \cdot 1 \, dx \).
We take \( u = \tan^{-1} x \) and \( v = 1 \). Using the rule for integration by parts, we get:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = \tan^{-1} x \int 1 \, dx - \int \left( \frac{d}{dx}(\tan^{-1} x) \int 1 \, dx \right) \, dx \)
\( = \tan^{-1} x \cdot x - \int \frac{1}{1+x^2} \cdot x \, dx \)
\( = x \tan^{-1} x - \int \frac{x}{1+x^2} \, dx \)
To solve \( \int \frac{x}{1+x^2} \, dx \), let \( t = 1+x^2 \). Then \( dt = 2x \, dx \), so \( x \, dx = \frac{1}{2} dt \).
\( \int \frac{x}{1+x^2} \, dx = \int \frac{1}{t} \cdot \frac{1}{2} dt = \frac{1}{2} \log |t| = \frac{1}{2} \log |1+x^2| \).
Therefore,
\( I = x \tan^{-1} x - \frac{1}{2} \log |1+x^2| + C \)
In simple words: To integrate \( \tan^{-1} x \), we treat it as \( \tan^{-1} x \cdot 1 \) and use integration by parts. We choose \( \tan^{-1} x \) as \( u \) and \( 1 \) as \( v \). The remaining integral \( \int \frac{x}{1+x^2} \, dx \) is solved by a simple substitution where \( 1+x^2 \) becomes \( t \).

Exam Tip: For inverse trigonometric functions without an explicit \( x \) term, multiply by 1 and use integration by parts with \( u = f^{-1}(x) \) and \( v = 1 \).

Question 14. \( x (\log x)^2 \)
Answer:
\( I = \int x (\log x)^2 \, dx \)
We take \( u = (\log x)^2 \) and \( v = x \). Applying the rule for integration by parts, we get:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = (\log x)^2 \int x \, dx - \int \left( \frac{d}{dx}((\log x)^2) \int x \, dx \right) \, dx \)
\( = (\log x)^2 \cdot \frac{x^2}{2} - \int \left( 2 \log x \cdot \frac{1}{x} \right) \cdot \frac{x^2}{2} \, dx \)
\( = \frac{x^2}{2} (\log x)^2 - \int x \log x \, dx \)
Now, we apply integration by parts again for \( \int x \log x \, dx \). We take \( u = \log x \) and \( v = x \):
\( I = \frac{x^2}{2} (\log x)^2 - \left[ \log x \int x \, dx - \int \left( \frac{d}{dx}(\log x) \int x \, dx \right) \, dx \right] \)
\( = \frac{x^2}{2} (\log x)^2 - \left[ \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx \right] \)
\( = \frac{x^2}{2} (\log x)^2 - \left[ \frac{x^2}{2} \log x - \int \frac{x}{2} \, dx \right] \)
\( = \frac{x^2}{2} (\log x)^2 - \left[ \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} \right] + C \)
\( = \frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C \)
In simple words: This integral needs the integration by parts rule to be applied two times. We start with \( u = (\log x)^2 \) and \( v = x \). After the first application, we get a new integral \( \int x \log x \, dx \), which we then solve again using integration by parts.

Exam Tip: When \( \log x \) is raised to a power, repeated integration by parts is often necessary. Always choose \( (\log x)^n \) as 'u' as its derivative simplifies.

Question 15. \( (x^2 + 1) \log x \)
Answer:
\( I = \int (x^2 + 1) \log x \, dx \)
We take \( u = \log x \) and \( v = x^2 + 1 \). Using the rule for integration by parts, we get:
\( \int uv \, dx = u \int v \, dx - \int \left( \frac{d}{dx}(u) \int v \, dx \right) \, dx \)
\( I = \log x \int (x^2 + 1) \, dx - \int \left( \frac{d}{dx}(\log x) \int (x^2 + 1) \, dx \right) \, dx \)
\( = \log x \left( \frac{x^3}{3} + x \right) - \int \frac{1}{x} \left( \frac{x^3}{3} + x \right) \, dx \)
\( = \left( \frac{x^3}{3} + x \right) \log x - \int \left( \frac{x^2}{3} + 1 \right) \, dx \)
\( = \left( \frac{x^3}{3} + x \right) \log x - \left( \frac{x^3}{3 \cdot 3} + x \right) + C \)
\( = \left( \frac{x^3}{3} + x \right) \log x - \frac{x^3}{9} - x + C \)
In simple words: To find this integral, we use the integration by parts method. We select \( \log x \) as \( u \) and \( x^2 + 1 \) as \( v \). After applying the formula, we integrate the resulting polynomial term to get the final solution.

Exam Tip: Remember to integrate each term in the polynomial \( (x^2+1) \) separately when finding \( \int v \, dx \).

Question 17. \( \frac{x e^x}{(1+x)^2} \)
Answer:
\( I = \int \frac{x e^x}{(1+x)^2} \, dx \)
We can rewrite the numerator as \( x e^x = (x+1-1) e^x \).
\( I = \int \frac{(x+1-1) e^x}{(1+x)^2} \, dx \)
\( = \int e^x \left[ \frac{x+1}{(1+x)^2} - \frac{1}{(1+x)^2} \right] \, dx \)
\( = \int e^x \left[ \frac{1}{1+x} - \frac{1}{(1+x)^2} \right] \, dx \)
This is in the form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \frac{1}{1+x} \).
Then \( f'(x) = \frac{d}{dx} (1+x)^{-1} = -1(1+x)^{-2} \cdot 1 = - \frac{1}{(1+x)^2} \).
So, using the formula, we get:
\( I = e^x \cdot \frac{1}{1+x} + C \)
In simple words: We modify the given expression to fit a special integral form: \( \int e^x [f(x) + f'(x)] \, dx \). We adjust the numerator to break the fraction into two parts. Then, we identify \( f(x) \) and its derivative \( f'(x) \) to directly apply the formula.

Exam Tip: Look for the pattern \( \int e^x [f(x) + f'(x)] \, dx \) when \( e^x \) is multiplied by a complex rational function. Strategic manipulation of the numerator is key.

Question 19. \( e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \)
Answer:
\( I = \int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx \)
This integral is in the form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \frac{1}{x} \).
Then, \( f'(x) = \frac{d}{dx} (x^{-1}) = -1x^{-2} = - \frac{1}{x^2} \).
Since the integrand is in the required form, we can directly apply the formula:
\( I = e^x \cdot \frac{1}{x} + C \)
In simple words: This integral exactly matches the special form \( \int e^x [f(x) + f'(x)] \, dx \). We can easily see that \( f(x) \) is \( \frac{1}{x} \) and its derivative \( f'(x) \) is \( - \frac{1}{x^2} \). So, we just use the formula to get the answer.

Exam Tip: Recognizing the \( \int e^x [f(x) + f'(x)] \, dx \) pattern is crucial for quickly solving certain integrals. Practice identifying \( f(x) \) and its derivative within the expression.

Question 20. \( \frac{(x-3) e^x}{(x-1)^3} \)
Answer:
\( I = \int \frac{(x-3) e^x}{(x-1)^3} \, dx \)
We can rewrite the numerator \( (x-3) \) as \( (x-1-2) \).
\( I = \int \frac{(x-1-2) e^x}{(x-1)^3} \, dx \)
\( = \int e^x \left[ \frac{x-1}{(x-1)^3} - \frac{2}{(x-1)^3} \right] \, dx \)
\( = \int e^x \left[ \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3} \right] \, dx \)
This is in the form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).
Here, let \( f(x) = \frac{1}{(x-1)^2} = (x-1)^{-2} \).
Then \( f'(x) = \frac{d}{dx} (x-1)^{-2} = -2(x-1)^{-3} \cdot 1 = - \frac{2}{(x-1)^3} \).
So, using the formula, we get:
\( I = e^x \cdot \frac{1}{(x-1)^2} + C \)
In simple words: To integrate this, we adjust the numerator \( (x-3) \) to \( (x-1-2) \) to split the fraction. This makes the expression fit the special form \( \int e^x [f(x) + f'(x)] \, dx \). We then identify \( f(x) \) and its derivative to find the solution.

Exam Tip: Complex rational functions with \( e^x \) often simplify into the \( e^x [f(x) + f'(x)] \) form. Look for algebraic manipulation that creates a function and its derivative.

Question 21. \( e^{2x} \sin x \)
Answer:
\( I = \int e^{2x} \sin x \, dx \)
We use integration by parts, taking \( u = \sin x \) and \( v = e^{2x} \).
\( I = \sin x \int e^{2x} \, dx - \int \left( \frac{d}{dx}(\sin x) \int e^{2x} \, dx \right) \, dx \)
\( = \sin x \cdot \frac{e^{2x}}{2} - \int \cos x \cdot \frac{e^{2x}}{2} \, dx \)
\( = \frac{e^{2x}}{2} \sin x - \frac{1}{2} \int e^{2x} \cos x \, dx \)
Now, we apply integration by parts again to \( \int e^{2x} \cos x \, dx \). We take \( u = \cos x \) and \( v = e^{2x} \):
\( I = \frac{e^{2x}}{2} \sin x - \frac{1}{2} \left[ \cos x \int e^{2x} \, dx - \int \left( \frac{d}{dx}(\cos x) \int e^{2x} \, dx \right) \, dx \right] \)
\( = \frac{e^{2x}}{2} \sin x - \frac{1}{2} \left[ \cos x \cdot \frac{e^{2x}}{2} - \int (-\sin x) \cdot \frac{e^{2x}}{2} \, dx \right] \)
\( = \frac{e^{2x}}{2} \sin x - \frac{1}{2} \left[ \frac{e^{2x}}{2} \cos x + \frac{1}{2} \int e^{2x} \sin x \, dx \right] \)
\( = \frac{e^{2x}}{2} \sin x - \frac{e^{2x}}{4} \cos x - \frac{1}{4} \int e^{2x} \sin x \, dx \)
Notice that \( \int e^{2x} \sin x \, dx \) is our original integral \( I \).
So, \( I = \frac{e^{2x}}{2} \sin x - \frac{e^{2x}}{4} \cos x - \frac{1}{4} I \)
\( I + \frac{1}{4} I = \frac{e^{2x}}{4} (2 \sin x - \cos x) \)
\( \frac{5}{4} I = \frac{e^{2x}}{4} (2 \sin x - \cos x) \)
\( I = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C \)
*Note: Remember the following formulas:*
(1) \( \int e^{ax} \sin(bx+k) \, dx = \frac{e^{ax}}{a^2+b^2} [a \sin(bx+k) - b \cos(bx+k)] + C \) (where \( a, b \neq 0 \))
(2) \( \int e^{ax} \cos(bx+k) \, dx = \frac{e^{ax}}{a^2+b^2} [a \cos(bx+k) + b \sin(bx+k)] + C \) (where \( a, b \neq 0 \))
For \( \int e^{2x} \sin x \, dx \), we have \( a=2 \), \( b=1 \), \( k=0 \).
So, \( \int e^{2x} \sin x \, dx = \frac{e^{2x}}{2^2+1^2} [2 \sin x - 1 \cos x] + C = \frac{e^{2x}}{5} [2 \sin x - \cos x] + C \).
In simple words: This integral needs integration by parts applied two times, as it results in the original integral appearing on the right side. We then solve for \( I \). Alternatively, one can use the direct formula for integrating \( e^{ax} \sin(bx) \) by identifying \( a \) and \( b \).

Exam Tip: For integrals of the form \( \int e^{ax} \sin(bx) \, dx \) or \( \int e^{ax} \cos(bx) \, dx \), apply integration by parts twice. The original integral will reappear, allowing you to solve for it algebraically. You can also memorize the direct formulas for these types of integrals.

Question 22. \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)
Answer:
\( I = \int \sin^{-1}\left(\frac{2x}{1+x^2}\right) \, dx \)
Let \( x = \tan t \). Then \( dx = \sec^2 t \, dt \).
The expression inside the inverse sine becomes \( \frac{2 \tan t}{1+\tan^2 t} = \frac{2 \tan t}{\sec^2 t} = 2 \sin t \cos t = \sin 2t \).
So, \( I = \int \sin^{-1}(\sin 2t) \sec^2 t \, dt \)
\( = \int 2t \sec^2 t \, dt \)
Now, we apply integration by parts to \( \int 2t \sec^2 t \, dt \). We take \( u = 2t \) and \( v = \sec^2 t \):
\( \int uv \, dt = u \int v \, dt - \int \left( \frac{d}{dt}(u) \int v \, dt \right) \, dt \)
\( I = 2t \int \sec^2 t \, dt - \int \left( \frac{d}{dt}(2t) \int \sec^2 t \, dt \right) \, dt \)
\( = 2t \tan t - \int 2 \tan t \, dt \)
\( = 2t \tan t - 2 \int \tan t \, dt \)
\( = 2t \tan t - 2 (-\log |\cos t|) + C \)
\( = 2t \tan t + 2 \log |\cos t| + C \)
Substitute back \( t = \tan^{-1} x \) and \( \tan t = x \). Since \( \tan t = x = \frac{x}{1} \), we can form a right triangle where the opposite side is \( x \) and the adjacent side is \( 1 \). The hypotenuse is \( \sqrt{x^2+1} \). Thus, \( \cos t = \frac{1}{\sqrt{x^2+1}} \).
So, \( I = 2 (\tan^{-1} x) x + 2 \log \left| \frac{1}{\sqrt{x^2+1}} \right| + C \)
\( = 2x \tan^{-1} x + 2 \log ( (x^2+1)^{-1/2} ) + C \)
\( = 2x \tan^{-1} x + 2 \left( -\frac{1}{2} \right) \log |x^2+1| + C \)
\( = 2x \tan^{-1} x - \log |x^2+1| + C \)
In simple words: First, we use the substitution \( x = \tan t \) to simplify the inverse sine term, making it \( \sin 2t \). Then, the integral becomes \( \int 2t \sec^2 t \, dt \). We solve this using integration by parts. Finally, we convert all \( t \) terms back to \( x \) using the original substitution and properties of logarithms.

Exam Tip: When an inverse trigonometric function has an argument like \( \frac{2x}{1+x^2} \), \( \frac{1-x^2}{1+x^2} \), or \( \frac{2x}{1-x^2} \), a substitution of \( x = \tan \theta \) (or \( x = \sin \theta \)) often simplifies it into a standard trigonometric identity.

Question 23. \( \int x^2 e^{x^3} dx = \)
(a) \( \frac{1}{3}e^{x^3} + c \)
(b) \( \frac{1}{3}e^{x^2} + c \)
(c) \( \frac{1}{2}e^{x^3} + c \)
(d) \( \frac{1}{2}e^{x^2} + c \)
Answer: (a) \( \frac{1}{3}e^{x^3} + c \)
In simple words: આપણે \( x^3 = t \) ધારીએ છીએ, તેથી \( 3x^2 dx = dt \) થાય. આથી, \( x^2 dx = \frac{dt}{3} \) બને છે. હવે સંકલન કરતા, આપણને \( I = \int e^t \frac{dt}{3} = \frac{1}{3}e^t + c \) મળે છે. અંતે, \( t \) ની જગ્યાએ પાછું \( x^3 \) મૂકવાથી, સાચો જવાબ \( \frac{1}{3}e^{x^3} + c \) મળે છે.

Exam Tip: For integration by substitution, always correctly find the derivative of the substituted term and adjust the differential \( dx \) accordingly.

Question 24. \( \int e^x \sec x(1 + \tan x) dx = \)
(a) \( e^x \cos x + c \)
(b) \( e^x \sec x + c \)
(c) \( e^x \sin x + c \)
(d) \( e^x \tan x + c \)
Answer: (b) \( e^x \sec x + c \)
In simple words: જો આપણે \( f(x) = \sec x \) તરીકે લઈએ, તો તેનું વિકલન \( f'(x) = \sec x \tan x \) થાય. આથી, આ પ્રશ્ન \( \int e^x [f(x) + f'(x)] dx \) ના સ્વરૂપમાં છે, જેનો ઉકેલ \( e^x f(x) + c \) છે. આ સૂત્રનો ઉપયોગ કરીને, આપણે \( e^x \sec x + c \) જવાબ તરીકે મેળવીએ છીએ.

Exam Tip: Recognize the integral form \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + c \) to quickly solve such problems. Identifying \( f(x) \) and \( f'(x) \) is key.