CBSE Class 12 Mathematics Integration Worksheet Set A - Free Printable

Let's solve each of the integrals step by step, explaining the methods used.

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Part 1: Basic Integrals

#### 1. $\int \sec^2(2 - 3x)\,dx$

Let $ u = 2 - 3x $. Then $ du = -3\,dx $ → $ dx = -\frac{1}{3}du $

$$
\int \sec^2(2 - 3x)\,dx = \int \sec^2(u) \cdot \left(-\frac{1}{3}\right)du = -\frac{1}{3} \tan u + C = -\frac{1}{3} \tan(2 - 3x) + C
$$

✔ Answer: $-\frac{1}{3} \tan(2 - 3x) + C$

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#### 2. $\int \frac{\sin(2 - ex)}{\cos^2(2 - 3x)}\,dx$

Wait — there’s a mismatch in variables: numerator has $ e $ (possibly Euler's number), but denominator uses $ x $. Let's assume it's a typo and should be:

> Likely meant: $\int \frac{\sin(2 - 3x)}{\cos^2(2 - 3x)}\,dx$

Let $ u = 2 - 3x $, so $ du = -3\,dx $, $ dx = -\frac{1}{3}du $

Then:
$$
\int \frac{\sin u}{\cos^2 u} \cdot \left(-\frac{1}{3}\right) du = -\frac{1}{3} \int \frac{\sin u}{\cos^2 u}\,du
$$

Let $ v = \cos u $, $ dv = -\sin u\,du $

So:
$$
-\frac{1}{3} \int \frac{-dv}{v^2} = \frac{1}{3} \int v^{-2} dv = \frac{1}{3} \left( -v^{-1} \right) + C = -\frac{1}{3v} + C = -\frac{1}{3\cos u} + C
$$

Substitute back:
$$
= -\frac{1}{3\cos(2 - 3x)} + C
$$

✔ Answer: $-\frac{1}{3\cos(2 - 3x)} + C$

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#### 3. $\int e^{2x - 3}\,dx$

Let $ u = 2x - 3 $, $ du = 2\,dx $, $ dx = \frac{1}{2}du $

$$
\int e^{2x - 3}\,dx = \int e^u \cdot \frac{1}{2} du = \frac{1}{2} e^u + C = \frac{1}{2} e^{2x - 3} + C
$$

✔ Answer: $\frac{1}{2} e^{2x - 3} + C$

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#### 4. $\int \sec(2 - 3x)\,dx$

Let $ u = 2 - 3x $, $ du = -3\,dx $, $ dx = -\frac{1}{3}du $

$$
\int \sec u \cdot \left(-\frac{1}{3}\right) du = -\frac{1}{3} \int \sec u\,du = -\frac{1}{3} \ln|\sec u + \tan u| + C
$$

Substitute back:
$$
= -\frac{1}{3} \ln|\sec(2 - 3x) + \tan(2 - 3x)| + C
$$

✔ Answer: $-\frac{1}{3} \ln|\sec(2 - 3x) + \tan(2 - 3x)| + C$

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#### 5. $\int \frac{1}{\sqrt{4x + 1}}\,dx$

Let $ u = 4x + 1 $, $ du = 4\,dx $, $ dx = \frac{1}{4}du $

$$
\int \frac{1}{\sqrt{u}} \cdot \frac{1}{4} du = \frac{1}{4} \int u^{-1/2} du = \frac{1}{4} \cdot 2u^{1/2} + C = \frac{1}{2} \sqrt{4x + 1} + C
$$

✔ Answer: $\frac{1}{2} \sqrt{4x + 1} + C$

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Part 2: Express numerator in terms of denominator

We are to express numerator in terms of denominator for these integrals.

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(i) $\int \frac{x - 1}{x + 1}\,dx$

Rewrite numerator:
$$
x - 1 = (x + 1) - 2
\Rightarrow \frac{x - 1}{x + 1} = \frac{(x + 1) - 2}{x + 1} = 1 - \frac{2}{x + 1}
$$

So:
$$
\int \left(1 - \frac{2}{x + 1}\right) dx = x - 2\ln|x + 1| + C
$$

✔ Answer: $x - 2\ln|x + 1| + C$

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(ii) $\int \frac{x^2 - 1}{x^2 + 1}\,dx$

Note: $x^2 - 1 = (x^2 + 1) - 2$, so:
$$
\frac{x^2 - 1}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}
$$

So:
$$
\int \left(1 - \frac{2}{x^2 + 1}\right) dx = x - 2\tan^{-1}x + C
$$

✔ Answer: $x - 2\tan^{-1}x + C$

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(iii) $\int \frac{x}{(2x + 1)^2}\,dx$

Let $ u = 2x + 1 $, then $ du = 2dx $, $ dx = \frac{1}{2}du $

Also, $ x = \frac{u - 1}{2} $

So:
$$
\int \frac{\frac{u - 1}{2}}{u^2} \cdot \frac{1}{2} du = \frac{1}{4} \int \frac{u - 1}{u^2} du = \frac{1}{4} \int \left(\frac{1}{u} - \frac{1}{u^2}\right) du
$$

$$
= \frac{1}{4} \left( \ln|u| + \frac{1}{u} \right) + C = \frac{1}{4} \left( \ln|2x + 1| + \frac{1}{2x + 1} \right) + C
$$

✔ Answer: $\frac{1}{4} \ln|2x + 1| + \frac{1}{4(2x + 1)} + C$

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(iv) $\int \frac{x^2 - 1}{x^2 + 1}\,dx$ ← Wait! This is same as (ii)

Already solved: $x - 2\tan^{-1}x + C$

But let's check: yes, identical to (ii). Possibly a typo?

Assuming it's correct, same answer.

✔ Answer: $x - 2\tan^{-1}x + C$

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(v) $\int \frac{x^2}{x + 1}\,dx$

Use polynomial division:

Divide $x^2$ by $x + 1$:

- $x^2 \div x = x$
- Multiply: $x(x + 1) = x^2 + x$
- Subtract: $x^2 - (x^2 + x) = -x$
- $-x \div x = -1$
- Multiply: $-1(x + 1) = -x - 1$
- Subtract: $-x - (-x - 1) = 1$

So:
$$
\frac{x^2}{x + 1} = x - 1 + \frac{1}{x + 1}
$$

Thus:
$$
\int \left(x - 1 + \frac{1}{x + 1}\right) dx = \frac{x^2}{2} - x + \ln|x + 1| + C
$$

✔ Answer: $\frac{x^2}{2} - x + \ln|x + 1| + C$

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(vi) $\int \frac{x^2}{(x + 1)^2}\,dx$

Let $ u = x + 1 $, so $ x = u - 1 $, $ dx = du $

Then $ x^2 = (u - 1)^2 = u^2 - 2u + 1 $

So:
$$
\int \frac{u^2 - 2u + 1}{u^2} du = \int \left(1 - \frac{2}{u} + \frac{1}{u^2}\right) du = u - 2\ln|u| - \frac{1}{u} + C
$$

Substitute back:
$$
= (x + 1) - 2\ln|x + 1| - \frac{1}{x + 1} + C
$$

✔ Answer: $x + 1 - 2\ln|x + 1| - \frac{1}{x + 1} + C$

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Part 3: Use hint $\int [f(x)]^n f'(x)\,dx = \frac{[f(x)]^{n+1}}{n+1} + C$

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(i) $\int \frac{(\sin^{-1}x)^2}{\sqrt{1 - x^2}}\,dx$

Let $ f(x) = \sin^{-1}x $, then $ f'(x) = \frac{1}{\sqrt{1 - x^2}} $

So the integral becomes:
$$
\int [f(x)]^2 \cdot f'(x)\,dx = \frac{[f(x)]^3}{3} + C = \frac{(\sin^{-1}x)^3}{3} + C
$$

✔ Answer: $\frac{(\sin^{-1}x)^3}{3} + C$

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(ii) $\int \sec^4 x \tan x\,dx$

Let $ f(x) = \sec x $, $ f'(x) = \sec x \tan x $

But we have $ \sec^4 x \tan x = \sec^3 x \cdot (\sec x \tan x) $

So write:
$$
\int \sec^3 x \cdot (\sec x \tan x)\,dx
$$

Let $ u = \sec x $, $ du = \sec x \tan x\,dx $

Then:
$$
\int u^3 du = \frac{u^4}{4} + C = \frac{\sec^4 x}{4} + C
$$

✔ Answer: $\frac{\sec^4 x}{4} + C$

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(iii) $\int \frac{\log^6 x}{x}\,dx$

Let $ f(x) = \log x $, $ f'(x) = \frac{1}{x} $

Then:
$$
\int [\log x]^6 \cdot \frac{1}{x}\,dx = \frac{[\log x]^7}{7} + C
$$

✔ Answer: $\frac{(\log x)^7}{7} + C$

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(iv) $\int \frac{x}{(x^2 + 1)^2}\,dx$

Let $ f(x) = x^2 + 1 $, $ f'(x) = 2x $

Then:
$$
\frac{x}{(x^2 + 1)^2} = \frac{1}{2} \cdot \frac{2x}{(x^2 + 1)^2} = \frac{1}{2} \cdot \frac{f'(x)}{[f(x)]^2}
$$

So:
$$
\int \frac{1}{2} \cdot \frac{f'(x)}{[f(x)]^2} dx = \frac{1}{2} \int [f(x)]^{-2} f'(x)\,dx = \frac{1}{2} \cdot \frac{[f(x)]^{-1}}{-1} + C = -\frac{1}{2(x^2 + 1)} + C
$$

✔ Answer: $-\frac{1}{2(x^2 + 1)} + C$

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(v) $\int \sin^5 x \cos x\,dx$

Let $ f(x) = \sin x $, $ f'(x) = \cos x $

Then:
$$
\int [\sin x]^5 \cdot \cos x\,dx = \frac{[\sin x]^6}{6} + C
$$

✔ Answer: $\frac{\sin^6 x}{6} + C$

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Part 4: Use hint $\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + C$

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(i) $\int \frac{x^3}{1 + x^4}\,dx$

Let $ f(x) = 1 + x^4 $, $ f'(x) = 4x^3 $

So:
$$
\frac{x^3}{1 + x^4} = \frac{1}{4} \cdot \frac{4x^3}{1 + x^4} = \frac{1}{4} \cdot \frac{f'(x)}{f(x)}
$$

So:
$$
\int \frac{x^3}{1 + x^4} dx = \frac{1}{4} \log|1 + x^4| + C
$$

Since $1 + x^4 > 0$, drop absolute value:

✔ Answer: $\frac{1}{4} \log(1 + x^4) + C$

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(ii) $\int \frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx$

Let $ f(x) = e^x + e^{-x} $, $ f'(x) = e^x - e^{-x} $

So:
$$
\int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C = \log(e^x + e^{-x}) + C
$$

✔ Answer: $\log(e^x + e^{-x}) + C$

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(iii) $\int \frac{e^x + 1}{e^x - 1}\,dx$

This is not directly in form $ \frac{f'}{f} $, but try manipulation:

Let $ u = e^x $, $ du = e^x dx $, $ dx = \frac{du}{u} $

Then:
$$
\int \frac{u + 1}{u - 1} \cdot \frac{du}{u} = \int \frac{u + 1}{u(u - 1)} du
$$

Partial fractions:

Let $ \frac{u + 1}{u(u - 1)} = \frac{A}{u} + \frac{B}{u - 1} $

Multiply:
$$
u + 1 = A(u - 1) + B u
$$

Set $ u = 0 $: $1 = -A$ → $A = -1$

Set $ u = 1 $: $2 = B$ → $B = 2$

So:
$$
\int \left( -\frac{1}{u} + \frac{2}{u - 1} \right) du = -\ln|u| + 2\ln|u - 1| + C = -\ln|e^x| + 2\ln|e^x - 1| + C
$$

$$
= -x + 2\ln(e^x - 1) + C
$$

✔ Answer: $-x + 2\ln(e^x - 1) + C$

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(iv) $\int \frac{\tan x + 1}{(\tan x + 1)^2}\,dx = \int \frac{1}{\tan x + 1}\,dx$

Wait — numerator and denominator both have $ \tan x + 1 $, so simplifies:

$$
\int \frac{1}{\tan x + 1}\,dx
$$

Let $ u = \tan x + 1 $, $ du = \sec^2 x\,dx $

But this doesn't help directly. Try substitution:

Alternatively, write:
$$
\int \frac{1}{\tan x + 1} dx = \int \frac{\cos x}{\sin x + \cos x} dx
$$

Let $ I = \int \frac{\cos x}{\sin x + \cos x} dx $

Let $ J = \int \frac{\sin x}{\sin x + \cos x} dx $

Then $ I + J = \int 1 dx = x + C $

And $ I - J = \int \frac{\cos x - \sin x}{\sin x + \cos x} dx = \int \frac{d(\sin x + \cos x)}{\sin x + \cos x} = \log|\sin x + \cos x| + C $

So:
$$
I + J = x, \quad I - J = \log|\sin x + \cos x|
$$

Add: $ 2I = x + \log|\sin x + \cos x| $ → $ I = \frac{1}{2}x + \frac{1}{2}\log|\sin x + \cos x| + C $

So:
$$
\int \frac{1}{\tan x + 1} dx = \frac{1}{2}x + \frac{1}{2}\log|\sin x + \cos x| + C
$$

✔ Answer: $\frac{1}{2}x + \frac{1}{2}\log|\sin x + \cos x| + C$

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(v) $\int \frac{\tan x + 1}{\tan x - 1}\,dx$

Write $ \frac{\tan x + 1}{\tan x - 1} = \frac{\frac{\sin x}{\cos x} + 1}{\frac{\sin x}{\cos x} - 1} = \frac{\sin x + \cos x}{\sin x - \cos x} $

Now let $ u = \sin x - \cos x $, $ du = (\cos x + \sin x) dx $

So:
$$
\int \frac{\sin x + \cos x}{\sin x - \cos x} dx = \int \frac{du}{u} = \log|u| + C = \log|\sin x - \cos x| + C
$$

✔ Answer: $\log|\sin x - \cos x| + C$

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(vi) $\int \frac{\sec x}{\log(\sec x - \tan x)}\,dx$

Let $ u = \log(\sec x - \tan x) $

Then compute $ du $:

Note that:
$$
\frac{d}{dx} (\sec x - \tan x) = \sec x \tan x - \sec^2 x = \sec x(\tan x - \sec x)
$$

So:
$$
\frac{d}{dx} \log(\sec x - \tan x) = \frac{1}{\sec x - \tan x} \cdot (\sec x \tan x - \sec^2 x) = \frac{\sec x(\tan x - \sec x)}{\sec x - \tan x} = -\sec x
$$

Because $ \tan x - \sec x = -(\sec x - \tan x) $

So:
$$
\frac{d}{dx} \log(\sec x - \tan x) = -\sec x
\Rightarrow d[\log(\sec x - \tan x)] = -\sec x\,dx
$$

Therefore:
$$
\int \frac{\sec x}{\log(\sec x - \tan x)} dx = \int \frac{-1}{u} du = -\log|u| + C = -\log|\log(\sec x - \tan x)| + C
$$

✔ Answer: $-\log|\log(\sec x - \tan x)| + C$

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✔ Final Summary of Answers:

#### Part 1:
1. $-\frac{1}{3} \tan(2 - 3x) + C$
2. $-\frac{1}{3\cos(2 - 3x)} + C$
3. $\frac{1}{2} e^{2x - 3} + C$
4. $-\frac{1}{3} \ln|\sec(2 - 3x) + \tan(2 - 3x)| + C$
5. $\frac{1}{2} \sqrt{4x + 1} + C$

#### Part 2:
(i) $x - 2\ln|x + 1| + C$
(ii) $x - 2\tan^{-1}x + C$
(iii) $\frac{1}{4} \ln|2x + 1| + \frac{1}{4(2x + 1)} + C$
(iv) $x - 2\tan^{-1}x + C$
(v) $\frac{x^2}{2} - x + \ln|x + 1| + C$
(vi) $x + 1 - 2\ln|x + 1| - \frac{1}{x + 1} + C$

#### Part 3:
(i) $\frac{(\sin^{-1}x)^3}{3} + C$
(ii) $\frac{\sec^4 x}{4} + C$
(iii) $\frac{(\log x)^7}{7} + C$
(iv) $-\frac{1}{2(x^2 + 1)} + C$
(v) $\frac{\sin^6 x}{6} + C$

#### Part 4:
(i) $\frac{1}{4} \log(1 + x^4) + C$
(ii) $\log(e^x + e^{-x}) + C$
(iii) $-x + 2\ln(e^x - 1) + C$
(iv) $\frac{1}{2}x + \frac{1}{2}\log|\sin x + \cos x| + C$
(v) $\log|\sin x - \cos x| + C$
(vi) $-\log|\log(\sec x - \tan x)| + C$

Let me know if you'd like any explanation visualized or further simplified!