Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.7

Integrate the Following with Respect to X:

Question 1.
(i) \( 9x e^{3x} \)
(ii) \( x \sin 3x \)
(iii) \( 25x e^{-5x} \)
(iv) \( x \sec x \tan x \)
Answer:
(i) To integrate \( 9x e^{3x} \) with respect to \( x \), we use integration by parts, which is given by the formula \( \int u \,dv = uv - \int v \,du \).
First, we simplify the expression:
\( \int 9x e^{3x} \,dx = 9 \int x e^{3x} \,dx \)
Now, we choose \( u \) and \( dv \):
Let \( u = x \)
Then \( du = dx \)
Let \( dv = e^{3x} \,dx \)
Then \( v = \int e^{3x} \,dx = \frac{e^{3x}}{3} \)
Using the integration by parts formula: \( \int x e^{3x} \,dx = x \left( \frac{e^{3x}}{3} \right) - \int \frac{e^{3x}}{3} \,dx \)
\( \implies \int x e^{3x} \,dx = \frac{x e^{3x}}{3} - \frac{1}{3} \int e^{3x} \,dx \)
\( \implies \int x e^{3x} \,dx = \frac{x e^{3x}}{3} - \frac{1}{3} \left( \frac{e^{3x}}{3} \right) + C \)
\( \implies \int x e^{3x} \,dx = \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} + C \)
Now, multiply by the constant 9 from the original integral:
\( 9 \int x e^{3x} \,dx = 9 \left( \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} \right) + C \)
\( \implies 9 \int x e^{3x} \,dx = 3x e^{3x} - e^{3x} + C \)
We can factor out \( e^{3x} \):
\( \implies 9 \int x e^{3x} \,dx = e^{3x} (3x - 1) + C \)
This method helps us find the integral of products of functions efficiently.
In simple words: To solve this, we use a special rule called integration by parts. We break the problem into two parts, find their derivatives and integrals, and then combine them using the formula. This helps us find the area under the curve of the given function.

(ii) To integrate \( x \sin 3x \) with respect to \( x \), we use the integration by parts formula: \( \int u \,dv = uv - u'v_1 + u''v_2 - u'''v_3 + \dots \)
Let \( u = x \)
Then \( u' = 1 \)
And \( u'' = 0 \)
Let \( dv = \sin 3x \,dx \)
Then \( v = \int \sin 3x \,dx = -\frac{\cos 3x}{3} \)
Next, \( v_1 = \int v \,dx = \int -\frac{\cos 3x}{3} \,dx = -\frac{1}{3} \int \cos 3x \,dx = -\frac{1}{3} \left( \frac{\sin 3x}{3} \right) = -\frac{\sin 3x}{9} \)
Using the integration by parts formula: \( \int x \sin 3x \,dx = x \left( -\frac{\cos 3x}{3} \right) - (1) \left( -\frac{\sin 3x}{9} \right) + (0) \left( \dots \right) + C \)
\( \implies \int x \sin 3x \,dx = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + C \)
Integration by parts is often used when one function becomes simpler by differentiation, and the other can be easily integrated.
In simple words: We used the integration by parts method. We picked one part to differentiate and another to integrate, then put them into a special formula. This helps us find the integral of a product of two functions.

(iii) To integrate \( 25x e^{-5x} \) with respect to \( x \), we use integration by parts, which is given by the formula \( \int u \,dv = uv - \int v \,du \).
First, we simplify the expression:
\( \int 25x e^{-5x} \,dx = 25 \int x e^{-5x} \,dx \)
Now, we choose \( u \) and \( dv \):
Let \( u = x \)
Then \( du = dx \)
Let \( dv = e^{-5x} \,dx \)
Then \( v = \int e^{-5x} \,dx = \frac{e^{-5x}}{-5} \)
Using the integration by parts formula: \( \int x e^{-5x} \,dx = x \left( \frac{e^{-5x}}{-5} \right) - \int \frac{e^{-5x}}{-5} \,dx \)
\( \implies \int x e^{-5x} \,dx = -\frac{x e^{-5x}}{5} + \frac{1}{5} \int e^{-5x} \,dx \)
\( \implies \int x e^{-5x} \,dx = -\frac{x e^{-5x}}{5} + \frac{1}{5} \left( \frac{e^{-5x}}{-5} \right) + C \)
\( \implies \int x e^{-5x} \,dx = -\frac{x e^{-5x}}{5} - \frac{e^{-5x}}{25} + C \)
Now, multiply by the constant 25 from the original integral:
\( 25 \int x e^{-5x} \,dx = 25 \left( -\frac{x e^{-5x}}{5} - \frac{e^{-5x}}{25} \right) + C \)
\( \implies 25 \int x e^{-5x} \,dx = -5x e^{-5x} - e^{-5x} + C \)
We can factor out \( e^{-5x} \):
\( \implies 25 \int x e^{-5x} \,dx = -e^{-5x} (5x + 1) + C \)
This method is particularly helpful for integrating products where one function simplifies upon repeated differentiation.
In simple words: We use integration by parts to solve this. We break the problem into parts, differentiate one and integrate the other, and then use a specific formula. This helps us find the integral of the given expression easily.

(iv) To integrate \( x \sec x \tan x \) with respect to \( x \), we use integration by parts, which is given by the formula \( \int u \,dv = uv - \int v \,du \).
First, we choose \( u \) and \( dv \):
Let \( u = x \)
Then \( du = dx \)
Let \( dv = \sec x \tan x \,dx \)
Then \( v = \int \sec x \tan x \,dx = \sec x \)
Using the integration by parts formula: \( \int x \sec x \tan x \,dx = x (\sec x) - \int \sec x \,dx \)
\( \implies \int x \sec x \tan x \,dx = x \sec x - \log |\sec x + \tan x| + C \)
This type of integral is common in advanced calculus, and knowing the standard integrals of trigonometric functions is key.
In simple words: We used integration by parts. We picked 'x' to differentiate and 'sec x tan x' to integrate. Then we put these into the formula, and remember that the integral of 'sec x' is 'log |sec x + tan x|'. This gives us the final answer.

🎯 Exam Tip: When using integration by parts, remember the "ILATE" rule to choose \( u \) (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) to simplify calculations. Also, correctly applying standard integral formulas for trigonometric functions is crucial.

Question 2.
(i) \( x \log x \)
(ii) \( 27x^2 e^{3x} \)
(iii) \( x^2 \cos x \)
(iv) \( x^3 \sin x \)
Answer:
(i) To integrate \( x \log x \) with respect to \( x \), we use integration by parts: \( \int u \,dv = uv - \int v \,du \).
According to the ILATE rule, we choose \( u = \log x \) because it is a logarithmic function, and \( dv = x \,dx \) because it is algebraic.
Let \( u = \log x \)
Then \( du = \frac{1}{x} \,dx \)
Let \( dv = x \,dx \)
Then \( v = \int x \,dx = \frac{x^2}{2} \)
Using the integration by parts formula: \( \int x \log x \,dx = (\log x) \left( \frac{x^2}{2} \right) - \int \frac{x^2}{2} \left( \frac{1}{x} \right) \,dx \)
\( \implies \int x \log x \,dx = \frac{x^2}{2} \log x - \int \frac{x}{2} \,dx \)
\( \implies \int x \log x \,dx = \frac{x^2}{2} \log x - \frac{1}{2} \int x \,dx \)
\( \implies \int x \log x \,dx = \frac{x^2}{2} \log x - \frac{1}{2} \left( \frac{x^2}{2} \right) + C \)
\( \implies \int x \log x \,dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + C \)
This is a fundamental example of using integration by parts to solve integrals involving logarithmic functions.
In simple words: We used integration by parts. We chose \( \log x \) to be differentiated and \( x \) to be integrated. Then we applied the formula, which helped us find the integral of this product.

(ii) To integrate \( 27x^2 e^{3x} \) with respect to \( x \), we use the generalized integration by parts formula: \( \int u \,dv = uv - u'v_1 + u''v_2 - u'''v_3 + \dots \)
First, we factor out the constant:
\( \int 27x^2 e^{3x} \,dx = 27 \int x^2 e^{3x} \,dx \)
Now, we choose \( u \) and \( dv \):
Let \( u = x^2 \)
Then \( u' = 2x \)
Then \( u'' = 2 \)
Then \( u''' = 0 \)
Let \( dv = e^{3x} \,dx \)
Then \( v = \int e^{3x} \,dx = \frac{e^{3x}}{3} \)
Then \( v_1 = \int v \,dx = \int \frac{e^{3x}}{3} \,dx = \frac{1}{3} \int e^{3x} \,dx = \frac{1}{3} \left( \frac{e^{3x}}{3} \right) = \frac{e^{3x}}{9} \)
Then \( v_2 = \int v_1 \,dx = \int \frac{e^{3x}}{9} \,dx = \frac{1}{9} \int e^{3x} \,dx = \frac{1}{9} \left( \frac{e^{3x}}{3} \right) = \frac{e^{3x}}{27} \)
Now, apply the formula to \( \int x^2 e^{3x} \,dx \):
\( \int x^2 e^{3x} \,dx = (x^2) \left( \frac{e^{3x}}{3} \right) - (2x) \left( \frac{e^{3x}}{9} \right) + (2) \left( \frac{e^{3x}}{27} \right) - (0) (\dots) + C \)
\( \implies \int x^2 e^{3x} \,dx = \frac{x^2 e^{3x}}{3} - \frac{2x e^{3x}}{9} + \frac{2 e^{3x}}{27} + C \)
Now, multiply by 27:
\( 27 \int x^2 e^{3x} \,dx = 27 \left( \frac{x^2 e^{3x}}{3} - \frac{2x e^{3x}}{9} + \frac{2 e^{3x}}{27} \right) + C \)
\( \implies 27 \int x^2 e^{3x} \,dx = 9x^2 e^{3x} - 6x e^{3x} + 2 e^{3x} + C \)
We can factor out \( e^{3x} \):
\( \implies 27 \int x^2 e^{3x} \,dx = e^{3x} (9x^2 - 6x + 2) + C \)
The generalized integration by parts is very useful when one function eventually differentiates to zero.
In simple words: We used a longer version of integration by parts because we had \( x^2 \). We differentiated \( x^2 \) until it became zero and integrated \( e^{3x} \) repeatedly. Then we put all these parts into the formula to get the final answer.

(iii) To integrate \( x^2 \cos x \) with respect to \( x \), we use the generalized integration by parts formula: \( \int u \,dv = uv - u'v_1 + u''v_2 - u'''v_3 + \dots \)
Now, we choose \( u \) and \( dv \):
Let \( u = x^2 \)
Then \( u' = 2x \)
Then \( u'' = 2 \)
Then \( u''' = 0 \)
Let \( dv = \cos x \,dx \)
Then \( v = \int \cos x \,dx = \sin x \)
Then \( v_1 = \int v \,dx = \int \sin x \,dx = -\cos x \)
Then \( v_2 = \int v_1 \,dx = \int -\cos x \,dx = -\sin x \)
Now, apply the formula to \( \int x^2 \cos x \,dx \):
\( \int x^2 \cos x \,dx = (x^2) (\sin x) - (2x) (-\cos x) + (2) (-\sin x) - (0) (\dots) + C \)
\( \implies \int x^2 \cos x \,dx = x^2 \sin x + 2x \cos x - 2 \sin x + C \)
This method makes integrating products of polynomials and trigonometric functions straightforward.
In simple words: We used the expanded integration by parts method. We kept differentiating \( x^2 \) and integrating \( \cos x \). Then we put all the results into the special formula. This helped us find the total integral.

(iv) To integrate \( x^3 \sin x \) with respect to \( x \), we use the generalized integration by parts formula: \( \int u \,dv = uv - u'v_1 + u''v_2 - u'''v_3 + u''''v_4 - \dots \)
Now, we choose \( u \) and \( dv \):
Let \( u = x^3 \)
Then \( u' = 3x^2 \)
Then \( u'' = 6x \)
Then \( u''' = 6 \)
Then \( u'''' = 0 \)
Let \( dv = \sin x \,dx \)
Then \( v = \int \sin x \,dx = -\cos x \)
Then \( v_1 = \int v \,dx = \int -\cos x \,dx = -\sin x \)
Then \( v_2 = \int v_1 \,dx = \int -\sin x \,dx = \cos x \)
Then \( v_3 = \int v_2 \,dx = \int \cos x \,dx = \sin x \)
Then \( v_4 = \int v_3 \,dx = \int \sin x \,dx = -\cos x \)
Now, apply the formula to \( \int x^3 \sin x \,dx \):
\( \int x^3 \sin x \,dx = (x^3) (-\cos x) - (3x^2) (-\sin x) + (6x) (\cos x) - (6) (\sin x) + (0) (-\cos x) + C \)
\( \implies \int x^3 \sin x \,dx = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C \)
This repeated application of integration by parts is effective for higher powers of polynomials.
In simple words: We used a long form of integration by parts. We differentiated \( x^3 \) multiple times until it became zero and integrated \( \sin x \) multiple times. Then we put these results into the formula to find the complete integral.

🎯 Exam Tip: When using the generalized integration by parts, carefully track the signs \( (+,-,+,-, \dots) \) and ensure each derivative of \( u \) is matched with the corresponding integral of \( dv \).

Question 3.
(i) \( \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \)
(ii) \( x^5 e^{x^2} \)
(iii) \( \tan^{-1} \left( \frac{8x}{1 - 16x^2} \right) \)
(iv) \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \)
Answer:
(i) To integrate \( \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \) with respect to \( x \), we use a substitution and then integration by parts.
Let \( \sin^{-1} x = \theta \).
Then \( x = \sin \theta \).
Then \( dx = \cos \theta \,d\theta \).
Also, \( \sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta \).
Substitute these into the integral:
\( \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \,dx = \int \frac{\sin \theta \cdot \theta}{\cos \theta} (\cos \theta \,d\theta) \)
\( \implies \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \,dx = \int \theta \sin \theta \,d\theta \)
Now, we use integration by parts for \( \int \theta \sin \theta \,d\theta \).
Let \( u = \theta \)
Then \( du = d\theta \)
Let \( dv = \sin \theta \,d\theta \)
Then \( v = \int \sin \theta \,d\theta = -\cos \theta \)
Using the integration by parts formula: \( \int \theta \sin \theta \,d\theta = \theta (-\cos \theta) - \int (-\cos \theta) \,d\theta \)
\( \implies \int \theta \sin \theta \,d\theta = -\theta \cos \theta + \int \cos \theta \,d\theta \)
\( \implies \int \theta \sin \theta \,d\theta = -\theta \cos \theta + \sin \theta + C \)
Now, substitute back \( \theta = \sin^{-1} x \) and \( \sin \theta = x \).
Since \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2} \), we have:
\( \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \,dx = -(\sin^{-1} x) \sqrt{1 - x^2} + x + C \)
This problem combines both substitution and integration by parts effectively.
In simple words: We first changed \( \sin^{-1} x \) to \( \theta \), which helped simplify the integral. Then, we used the integration by parts rule to solve the new integral. Finally, we changed \( \theta \) back to \( \sin^{-1} x \) to get the answer in terms of \( x \).

🎯 Exam Tip: For integrals involving inverse trigonometric functions with square roots, a trigonometric substitution (like \( x = \sin \theta \) or \( x = \tan \theta \)) often simplifies the expression before applying integration by parts.

Question 3.
(ii) \( x^5 e^{x^2} \)
Answer:
To integrate \( x^5 e^{x^2} \) with respect to \( x \), we use a substitution and then integration by parts.
Let \( t = x^2 \).
Then \( dt = 2x \,dx \).
\( \implies x \,dx = \frac{1}{2} \,dt \).
We can rewrite \( x^5 = x^4 \cdot x = (x^2)^2 \cdot x = t^2 \cdot x \).
Substitute these into the integral:
\( \int x^5 e^{x^2} \,dx = \int (x^2)^2 e^{x^2} (x \,dx) \)
\( \implies \int x^5 e^{x^2} \,dx = \int t^2 e^t \left( \frac{1}{2} \,dt \right) \)
\( \implies \int x^5 e^{x^2} \,dx = \frac{1}{2} \int t^2 e^t \,dt \)
Now, we use the generalized integration by parts formula for \( \int t^2 e^t \,dt \): \( \int u \,dv = uv - u'v_1 + u''v_2 - u'''v_3 + \dots \)
Let \( u = t^2 \)
Then \( u' = 2t \)
Then \( u'' = 2 \)
Then \( u''' = 0 \)
Let \( dv = e^t \,dt \)
Then \( v = \int e^t \,dt = e^t \)
Then \( v_1 = \int v \,dt = \int e^t \,dt = e^t \)
Then \( v_2 = \int v_1 \,dt = \int e^t \,dt = e^t \)
Apply the formula to \( \int t^2 e^t \,dt \):
\( \int t^2 e^t \,dt = (t^2) (e^t) - (2t) (e^t) + (2) (e^t) - (0) (\dots) + C \)
\( \implies \int t^2 e^t \,dt = t^2 e^t - 2t e^t + 2 e^t + C \)
\( \implies \int t^2 e^t \,dt = e^t (t^2 - 2t + 2) + C \)
Now, substitute back \( t = x^2 \):
\( \frac{1}{2} \int t^2 e^t \,dt = \frac{1}{2} e^{x^2} ((x^2)^2 - 2(x^2) + 2) + C \)
\( \implies \int x^5 e^{x^2} \,dx = \frac{1}{2} e^{x^2} (x^4 - 2x^2 + 2) + C \)
This example shows how combining substitution with integration by parts can solve complex integrals.
In simple words: First, we used a substitution by letting \( t = x^2 \), which made the integral simpler. Then, we applied the integration by parts rule repeatedly to solve the new integral. Finally, we replaced \( t \) with \( x^2 \) to get the answer in terms of \( x \).

🎯 Exam Tip: When integrating functions like \( x^n e^{ax^k} \), consider a substitution like \( t = x^k \) to simplify the exponent and then use integration by parts.

Question 3.
(iii) \( \tan^{-1} \left( \frac{8x}{1 - 16x^2} \right) \)
Answer:
To integrate \( \tan^{-1} \left( \frac{8x}{1 - 16x^2} \right) \) with respect to \( x \), we use a trigonometric substitution and then integration by parts.
We know the identity \( \tan^{-1} \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right) = \tan^{-1} (\tan 2\theta) = 2\theta \).
Let \( 4x = \tan \theta \).
Then \( 8x = 2 \tan \theta \).
And \( 16x^2 = \tan^2 \theta \).
Substitute these into the function:
\( \tan^{-1} \left( \frac{8x}{1 - 16x^2} \right) = \tan^{-1} \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right) = 2\theta \)
From \( 4x = \tan \theta \), we have \( \theta = \tan^{-1} (4x) \).
So, the integral becomes \( \int 2\theta \,dx \). However, we need to express \( dx \) in terms of \( d\theta \).
From \( 4x = \tan \theta \), differentiate with respect to \( \theta \):
\( 4 \,dx = \sec^2 \theta \,d\theta \)
\( \implies \,dx = \frac{1}{4} \sec^2 \theta \,d\theta \)
Now, substitute into the integral:
\( \int \tan^{-1} \left( \frac{8x}{1 - 16x^2} \right) \,dx = \int (2\theta) \left( \frac{1}{4} \sec^2 \theta \,d\theta \right) \)
\( \implies \int \tan^{-1} \left( \frac{8x}{1 - 16x^2} \right) \,dx = \frac{1}{2} \int \theta \sec^2 \theta \,d\theta \)
Now, we use integration by parts for \( \int \theta \sec^2 \theta \,d\theta \).
Let \( u = \theta \)
Then \( du = d\theta \)
Let \( dv = \sec^2 \theta \,d\theta \)
Then \( v = \int \sec^2 \theta \,d\theta = \tan \theta \)
Using the integration by parts formula: \( \int \theta \sec^2 \theta \,d\theta = \theta \tan \theta - \int \tan \theta \,d\theta \)
\( \implies \int \theta \sec^2 \theta \,d\theta = \theta \tan \theta - \log |\sec \theta| + C \)
Now, substitute back \( \theta = \tan^{-1} (4x) \) and \( \tan \theta = 4x \).
Also, \( \sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + (4x)^2} = \sqrt{1 + 16x^2} \).
Substitute these back into the expression for \( \int \theta \sec^2 \theta \,d\theta \):
\( \int \theta \sec^2 \theta \,d\theta = (\tan^{-1} (4x)) (4x) - \log |\sqrt{1 + 16x^2}| + C \)
Finally, multiply by \( \frac{1}{2} \):
\( \int \tan^{-1} \left( \frac{8x}{1 - 16x^2} \right) \,dx = \frac{1}{2} [4x \tan^{-1} (4x) - \log |\sqrt{1 + 16x^2}|] + C \)
This problem smartly uses a trigonometric identity to simplify the inverse tangent before integration.
In simple words: We used a special trick involving the tangent function to simplify the inside part of \( \tan^{-1} \). After that, we made a substitution and then used the integration by parts rule. Finally, we changed everything back to \( x \) to get the answer.

🎯 Exam Tip: Recognizing inverse trigonometric identities like \( \tan^{-1} \left( \frac{2x}{1-x^2} \right) = 2 \tan^{-1} x \) is key to simplifying these integrals. Always look for such patterns before attempting direct integration.

Question 3.
(iv) \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \)
Answer:
To integrate \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \) with respect to \( x \), we use a trigonometric substitution and then integration by parts.
We know the identity \( \sin^{-1} \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) = \sin^{-1} (\sin 2\theta) = 2\theta \).
Let \( x = \tan \theta \).
Then \( dx = \sec^2 \theta \,d\theta \).
Substitute these into the function:
\( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) = \sin^{-1} \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) = 2\theta \)
From \( x = \tan \theta \), we have \( \theta = \tan^{-1} x \).
So, the integral becomes:
\( \int \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \,dx = \int (2\theta) (\sec^2 \theta \,d\theta) \)
\( \implies \int \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \,dx = 2 \int \theta \sec^2 \theta \,d\theta \)
Now, we use integration by parts for \( \int \theta \sec^2 \theta \,d\theta \). (This is the same integral from part (iii))
Let \( u = \theta \)
Then \( du = d\theta \)
Let \( dv = \sec^2 \theta \,d\theta \)
Then \( v = \int \sec^2 \theta \,d\theta = \tan \theta \)
Using the integration by parts formula: \( \int \theta \sec^2 \theta \,d\theta = \theta \tan \theta - \int \tan \theta \,d\theta \)
\( \implies \int \theta \sec^2 \theta \,d\theta = \theta \tan \theta - \log |\sec \theta| + C' \)
Now, substitute back \( \theta = \tan^{-1} x \) and \( \tan \theta = x \).
Also, \( \sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + x^2} \).
Substitute these back into the expression for \( \int \theta \sec^2 \theta \,d\theta \):
\( \int \theta \sec^2 \theta \,d\theta = x \tan^{-1} x - \log |\sqrt{1 + x^2}| + C' \)
Finally, multiply by 2:
\( \int \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \,dx = 2 [x \tan^{-1} x - \log |\sqrt{1 + x^2}|] + C \)
This integral shows the power of trigonometric substitution to simplify inverse trigonometric functions before integrating.
In simple words: We first used the substitution \( x = \tan \theta \) to simplify the \( \sin^{-1} \) part, changing it into \( 2\theta \). Then, we used the integration by parts rule to solve the new integral. Lastly, we converted back from \( \theta \) to \( x \) to get the final answer.

🎯 Exam Tip: Always recognize the form \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \) as \( 2 \tan^{-1} x \) (if \( |x| \le 1 \)) or related identities, as this significantly simplifies the integration process.