Students can track their progress and improvement through regular use of ML Aggarwal Maths for Class 12 Solutions Chapter 8 Integrals Ex 8.7.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.7

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 16) integrals:

Question 1.
(i) ∫ $$\frac{1}{x^2-16}$$ dx (NCERT)
(ii) ∫ $$\frac{1}{9-x^2}$$ dx
(iii) ∫ $$\frac{d x}{\sqrt{1-x^2}}$$
(iv) ∫ $$\frac{d x}{x^2+16}$$
Solution:
(i) ∫ $$\frac{1}{x^2-16}$$ dx
= (i) ∫ $$\frac{d x}{x^2-16}$$
= $$\frac{1}{2 \times 4} \log \left|\frac{x-4}{x+4}\right|$$ + C
= $$\frac{1}{8} \log \left|\frac{x-4}{x+4}\right|$$ + C
[∵ ∫ $$\frac{d x}{x^2-a^2}$$ = $$\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|$$ + C]

(ii) ∫ $$\frac{1}{9-x^2}$$ dx
= ∫ $$\frac{d x}{3^2-x^2}$$
= $$\frac{1}{2 \times 3} \log \left|\frac{3+x}{3-x}\right|$$ + C
= $$\frac{1}{6} \log \left|\frac{3+x}{3-x}\right|$$ + C
[∵ ∫ $$\frac{d x}{x^2-a^2}$$ = $$\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|$$ + C]

(iii) ∫ $$\frac{d x}{\sqrt{1-x^2}}$$
= ∫ $$\frac{d x}{\sqrt{1^2-x^2}}$$
= sin-1 $$\frac{x}{1}$$ + C
= sin-1 x + C
[∵ ∫ $$\frac{d x}{a^2-x^2}$$ = sin-1 $$\frac{x}{a}$$ + C]

(iv) ∫ $$\frac{d x}{x^2+16}$$
= ∫ $$\frac{d x}{x^2+4^2}$$
=$$\frac{1}{4}$$ tan-1 $$\frac{x}{4}$$ + C
[∵ ∫ $$\frac{d x}{x^2+a^2}$$ = $$\frac{1}{a} \tan ^{-1} \frac{x}{a}$$ + C]

Question 2.
(i) ∫ $$\frac{d x}{\sqrt{9-x^2}}$$
(ii) ∫ $$\frac{d x}{\sqrt{25+9 x^2}}$$
Solution:
(i) Let I = ∫ $$\frac{d x}{\sqrt{9-x^2}}$$
= $$\frac{1}{2} \int \frac{d x}{\sqrt{\left(\frac{3}{2}\right)^2-x^2}}$$
= $$\frac{1}{2} \sin ^{-1}\left(\frac{x}{\frac{3}{2}}\right)$$ + C
[∵ ∫ $$\frac{d x}{x^2-a^2}$$ = sin-1 $$\frac{x}{a}$$ + C]
= $$\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{3}\right)$$ + C

(ii) ∫ $$\frac{d x}{\sqrt{25+9 x^2}}$$
= $$\frac{1}{9} \int \frac{d x}{x^2+\left(\frac{5}{3}\right)^2}$$
= $$\frac{1}{9} \times \frac{1}{\frac{5}{3}} \tan ^{-1}\left(\frac{x}{\frac{5}{3}}\right)$$ + C
= $$\frac{1}{15} \tan ^{-1}\left(\frac{3 x}{5}\right)$$ + C
[∵ ∫ $$\frac{d x}{x^2+a^2}$$ = $$\frac{1}{a} \tan ^{-1} \frac{x}{a}$$ + C]

Question 2 (old).
(i) ∫ $$\frac{d x}{\sqrt{16-9 x^2}}$$ (NCERT Exemplar)
Solution:
(i) ∫ $$\frac{d x}{\sqrt{16-9 x^2}}$$
= $$\frac{1}{3} \int \frac{d x}{\sqrt{\left(\frac{4}{3}\right)^2-x^2}}$$
= $$\frac{1}{3} \sin ^{-1}\left(\frac{x}{\frac{4}{3}}\right)$$ + C
[∵ ∫ $$\frac{d x}{a^2-x^2}$$ = sin-1 $$\frac{x}{a}$$ + C]
= $$\frac{1}{3} \sin ^{-1}\left(\frac{3 x}{4}\right)$$ + C

Question 3.
(i) ∫ $$\frac{d x}{x \sqrt{9 x^2-16}}$$
(ii) ∫ $$\frac{d x}{\sqrt{x^2+5}}$$
Solution:
(i) ∫ $$\frac{d x}{x \sqrt{9 x^2-16}}$$
= $$\frac{1}{3} \int \frac{d x}{x \sqrt{x^2-\left(\frac{4}{3}\right)^2}}$$
= $$\frac{1}{3} \times \frac{1}{\frac{4}{3}} \sec ^{-1}\left(\frac{x}{\frac{4}{3}}\right)$$ + C
= $$\frac{1}{4} \sec ^{-1}\left(\frac{3 x}{4}\right)$$ + C
[∵ ∫ $$\frac{d x}{x \sqrt{x^2-a^2}}=\frac{1}{a} \sec ^{-1} \frac{x}{a}$$ + C]

(ii) Let I = ∫ $$\frac{d x}{\sqrt{x^2+5}}$$
= ∫ $$\frac{d x}{\sqrt{x^2+(\sqrt{5})^2}}$$
= log |x + $$\sqrt{x^2+5}$$| + C
[∵ ∫ $$\frac{d x}{\sqrt{x^2+a^2}}$$ = log |x + $$\sqrt{x^2+a^2}$$| + C

Question 3 (old).
(ii) ∫ $$\frac{d x}{\sqrt{x^2+4}}$$
Solution:
∫ $$\frac{d x}{\sqrt{x^2+4}}$$
= ∫ $$\frac{d x}{\sqrt{x^2+2^2}}$$
= log |x + $$\sqrt{x^2+4}$$| + C
[∵ ∫ $$\frac{d x}{\sqrt{x^2+a^2}}$$ = log |x + $$\sqrt{x^2+a^2}$$| + C

Question 4.
(i) ∫ $$\frac{1}{9 x^2-1}$$ dx
(ii) ∫ $$\frac{d x}{32-2 x^2}$$
Solution:
(i) ∫ $$\frac{1}{9 x^2-1}$$ dx
= $$\frac{1}{9} \int \frac{d x}{x^2-\left(\frac{1}{3}\right)^2}$$
= $$\frac{1}{9} \times \frac{1}{2 \times \frac{1}{3}} \log \left|\frac{x-\frac{1}{3}}{x+\frac{1}{3}}\right|$$ + C
= $$\frac{1}{6} \log \left|\frac{3 x-1}{3 x+1}\right|$$ + C

(ii) ∫ $$\frac{d x}{32-2 x^2}$$
= $$\frac{1}{2} \int \frac{d x}{16-x^2}$$
= $$\frac{1}{2} \int \frac{d x}{4^2-x^2}$$
= $$\frac{1}{2} \times \frac{1}{2 \times 4} \log \left|\frac{4+x}{4-x}\right|$$ + C
= $$\frac{1}{16} \log \left|\frac{4+x}{4-x}\right|$$ + C
[∵ ∫ $$\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|$$ + C]

Question 5.
(i) ∫ $$\frac{d x}{\sqrt{15-8 x^2}}$$
(ii) ∫ $$\frac{d x}{\sqrt{1+4 x^2}}$$ (NCERT)
Solution:
(i) ∫ $$\frac{d x}{\sqrt{15-8 x^2}}$$

(ii) ∫ $$\frac{d x}{\sqrt{1+4 x^2}}$$
= $$\frac{1}{2} \int \frac{d x}{\sqrt{\left(\frac{1}{2}\right)^2+x^2}}$$
= $$\frac{1}{2} \log \left|x+\sqrt{x^2+\frac{1}{4}}\right|$$ + C
[∵ ∫ $$\frac{d x}{\sqrt{x^2+a^2}}$$ = log |x + $$\sqrt{x^2+a^2}$$| + C

Question 6.
(i) ∫ $$\frac{2 y^2}{y^2+4}$$ dy (ISC 2015)
(ii) ∫ $$\frac{x^4+1}{x^2+1}$$ dx
Solution:
(i) ∫ $$\frac{2 y^2}{y^2+4}$$ dy
= 2 ∫ $$\frac{y^2+4-4}{y^2+4}$$ dy
= 2 ∫ [1 – $$\frac{4}{y^2+4}$$] dy
= 2 ∫ dy – 8 ∫ $$\frac{d y}{y^2+2^2}$$
= 2y – $$\frac{8}{2} \tan ^{-1}\left(\frac{y}{2}\right)$$ + C
= 2y – 4 tan-1 ($$\frac{y}{2}$$) + C

(ii) Let I = ∫ $$\frac{x^4+1}{x^2+1}$$ dx
= ∫ [x2 – 1 + $$\frac{2}{x^2+1}$$] dx
= $$\frac{x^{3}}{3}$$ – x + 2 tan-1 $$\frac{x}{1}$$ + c
= $$\frac{x^{3}}{3}$$ – x + 2 tan-1 x + c
[∵ ∫ $$\frac{d x}{\sqrt{x^2+a^2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}$$ + c]

Question 7.
(i) ∫ $$\frac{x}{x^4-1}$$ dx (NCERT Exemplar)
(ii) ∫ $$\frac{3 x^2}{1+x^6}$$ dx (NCERT)
Solution:
(i) I = ∫ $$\frac{x}{x^4-1}$$ dx
put x2 = t
⇒ 2x dx = dt
∴ I = ∫ $$\int \frac{d t}{2\left(t^2-1\right)}$$
= $$\frac{1}{2} \int \frac{d t}{t^2-1^2}$$
= $$\frac{1}{2} \times \frac{1}{2 \times 1} \log \left|\frac{t-1}{t+1}\right|$$ + C
= $$\frac{1}{4} \log \left|\frac{x^2-1}{x^2+1}\right|$$ + C
[∵ ∫ $$\frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \mid \frac{x-a}{x+a}$$ + C]

(ii) Let I = $$\frac{3 x^2}{1+x^6}$$ dx
∴ I = ∫ $$\frac{d t}{t^2+1}$$
= tan-1 t + C
= tan-1 (x3) + C

Question 8.
(i) ∫ $$\frac{x^2}{\sqrt{x^6+a^6}}$$ dx (NCERT)
(ii) ∫ $$\frac{x^3}{\sqrt{1-x^8}}$$ dx (NCERT)
Solution:
(i) Let I = ∫ $$\frac{x^2}{\sqrt{x^6+a^6}}$$ dx
put x3 = t
⇒ 3x2 dx = dt
∴ I = ∫ $$\frac{d t}{3 \sqrt{t^2+\left(a^3\right)^2}}$$
= $$\frac{1}{3}$$ log |t + $$\sqrt{t^2+a^6}$$| + C
= $$\frac{1}{3}$$ log |x3 + $$\sqrt{x^6+a^6}$$| + C

(ii) I = ∫ $$\frac{x^3}{\sqrt{1-x^8}}$$ dx
put x4 = t
⇒ 4x3 dx = dt
∴ I = $$\int \frac{d t}{4 \sqrt{1-t^2}}$$
= $$\frac{1}{4} \sin ^{-1} \frac{t}{1}$$ + C
= $$\frac{1}{4}$$ sin-1 (x4) + C
[∵ ∫ $$\frac{d x}{\sqrt{a^2-x^2}}$$ = sin-1 $$\frac{x}{a}$$ + C]

Question 9.
(i) ∫ $$\frac{\sqrt{x}}{\sqrt{a^3-x^3}}$$ dx (NCERT Exemplar)
(ii) ∫ $$\frac{3 x}{1+2 x^4}$$ dx (NCERT)
Solution:
(i) Let I = ∫ $$\frac{\sqrt{x} d x}{\sqrt{a^3-x^3}}$$

(ii) Let I = ∫ $$\frac{3 x}{1+2 x^4}$$ dx
= ∫ $$\frac{3 x}{1+2\left(x^2\right)^2}$$ dx
put x2 = t
∴ 2x dx = dt
∴ I = 3 ∫ $$\frac{d t}{2\left[1+2 t^2\right]}$$
= $$\frac{3}{4} \int \frac{d t}{t^2+\frac{1}{2}}$$
= $$\frac{3}{4} \int \frac{d t}{t^2+\left(\frac{1}{\sqrt{2}}\right)^2}$$
= $$\frac{3}{4} \cdot \frac{1}{\frac{1}{\sqrt{2}}} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)$$ + C
= $$\frac{3 \sqrt{2}}{4}$$ tan-1 (√2 x2) + C

Question 10.
(i) ∫ $$\frac{\cos x}{\sqrt{4-\sin ^2 x}}$$ dx (NCERT)
(ii) ∫ $$\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}$$ dx (NCERT)
Solution:
(i) Let I = ∫ $$\frac{\cos x}{\sqrt{4-\sin ^2 x}}$$ dx
put sin x = t
⇒ cos x dx = dt
= ∫ $$\frac{d t}{\sqrt{4-t^2}}$$
= ∫ $$\frac{d t}{\sqrt{2^2-t^2}}$$
= sin-1 ($$\frac{t}{2}$$) + c
= sin-1 ($$\left(\frac{\sin x}{2}\right)$$) + c

(ii) Let I = ∫ $$\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}$$ dx
put tan x = t
⇒ ∫ sec2 x dx = dt
= ∫ $$\frac{d t}{\sqrt{4+t^2}}$$
= ∫ $$\frac{d t}{\sqrt{t^2+2^2}}$$
= log |t + $$\sqrt{t^2+4}$$| + c
[∵ ∫ $$\frac{d x}{\sqrt{x^2+a^2}}$$ = log |x + $$\sqrt{x^2+a^2}$$| + c
= log |tan x + $$\sqrt{\tan ^2 x+4}$$| + c

Question 11.
(i) ∫ $$\frac{\sin x}{16-9 \cos ^2 x}$$ dx
(ii) ∫ $$\frac{\tan ^2 x \sec ^2 x}{1-\tan ^6 x}$$ dx
Solution:
(i) Let I = ∫ $$\frac{\sin x}{16-9 \cos ^2 x}$$ dx
put cos x = t
⇒ – sin x dx = dt

(ii) Let I = ∫ $$\frac{\tan ^2 x \sec ^2 x}{1-\tan ^6 x}$$ dx
put tan3 x = t
⇒ 3 tan2 x sec2 x dx = dt
∴ I = ∫ $$\frac{d t}{3\left(1-t^2\right)}$$
= $$\frac{1}{3} \times \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|$$ + C
[∵ $$\frac{d x}{a^2+x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|$$ + C]
= $$\frac{1}{6} \log \left|\frac{1+\tan ^3 x}{1-\tan ^3 x}\right|$$ + C

Question 11 (old).
(ii) ∫ latex]\frac{\ {cosec}^2 x}{1-\cot ^2 x}[/latex] dx
Solution:
Let I = ∫ latex]\frac{\ {cosec}^2 x}{1-\cot ^2 x}[/latex] dx
put cos x = t
⇒ – cosec2 x dx = dt
∴ I = ∫ $$\frac{-d t}{1-t^2}$$
= – $$\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|$$
= – $$\frac{1}{2} \log \left|\frac{1+\cot x}{1-\cot x}\right|$$ + C
[∵ $$\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|$$ + C]

Question 12.
(i) ∫ $$\frac{\sin x+\cos x}{\sqrt{\sin 2 x}}$$ dx
(ii) ∫ $$\frac{d x}{2 \sin ^2 x+5 \cos ^2 x}$$
Solution:
(i) Let I = ∫ $$\frac{\sin x+\cos x}{\sqrt{\sin 2 x}}$$ dx
put sin x – cos x = t ……….(1)
⇒ (cos x + sin x) dx = dt
On squaring (1) ; we have
sin2 x + cos2 x – 2 sin x cos x = t2
⇒ 1 – sin 2x = t2
⇒ sin 2x = 1 – t2
∴ I = ∫ $$\frac{d t}{\sqrt{1-t^2}}$$
= sin-1 (t) + C
= sin-1 (sin x – cos x) + C

(ii) Let I = ∫ $$\frac{d x}{2 \sin ^2 x+5 \cos ^2 x}$$
Divide Numerator and denominator by cos2 x ; we get
put tan x = t
⇒ sec2 x dx = dt
∴ I = ∫ $$\frac{d t}{2 t^2+5}$$
= $$\frac{1}{2}$$ ∫ $$\frac{d t}{t^2+\left(\sqrt{\frac{5}{2}}\right)^2}$$
= $$\frac{1}{2} \times \frac{1}{\sqrt{\frac{5}{2}}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{2}}}\right)$$ + C
[∵ ∫ $$\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}$$ + C]
= $$\frac{1}{\sqrt{10}} \tan ^{-1}\left(\sqrt{\frac{2}{5}} \tan x\right)$$ + C

Question 12 (old).
(ii) ∫ $$\frac{\sin x+\cos x}{9+16 \sin 2 x}$$ dx
Solution:
Let I = ∫ $$\frac{\sin x+\cos x}{9+16 \sin 2 x}$$ dx

Question 13.
(i) ∫ $$\frac{\cos x}{\cos 3 x}$$ dx
(ii) ∫ $$\frac{d x}{2-3 \cos 2 x}$$
Solution:
(i) Let I = ∫ $$\frac{\cos x}{\cos 3 x}$$ dx

(ii) Let I = ∫ $$\frac{d x}{2-3 \cos 2 x}$$
= ∫ $$\frac{d x}{2-3\left(1-2 \sin ^2 x\right)}$$
= ∫ $$\frac{d x}{-1+6 \sin ^2 x}$$
Divide numerator and denominator by cos2 x ; we have
= ∫ $$\frac{\sec ^2 x d x}{-\sec ^2 x+6 \tan ^2 x}$$
=∫ $$\frac{\sec ^2 x d x}{-1-\tan ^2 x+6 \tan ^2 x}$$

Question 13 (old).
(ii) ∫ $$\frac{d x}{11 \sin ^2 x+4 \cos ^2 x+5}$$
Solution:
Let I = ∫ $$\frac{d x}{11 \sin ^2 x+4 \cos ^2 x+5}$$
Divide numerator and denominator by cos2 x ; we have

Question 14.
(i) ∫ $$\frac{d x}{e^x-e^{-x}}$$
(ii) ∫ $$\sqrt{\frac{1+x}{1-x}}$$ dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ $$\frac{d x}{e^x-e^{-x}}$$
= ∫ $$\frac{d x}{e^x-\frac{1}{e^x}}$$
= ∫ $$\frac{e^x d x}{e^{2 x}-1}$$
put ex = t
⇒ ex dx = dt
= ∫ $$\frac{d t}{t^2-1^2}$$
= $$\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|$$ + C
= $$\frac{1}{2} \log \left|\frac{e^x-1}{e^x+1}\right|$$ + C

(ii) Let I = ∫ $$\sqrt{\frac{1+x}{1-x}}$$ dx
put x = cos θ
⇒ dx = – sin θ dθ

= – [θ + sin θ] + C
= – cos-1 x – $$\sqrt{1-x^{2}}$$ + C
[∵ cos θ = x
⇒ θ = cos-1 x
and sin θ = $$\sqrt{1-\cos ^2 \theta}$$
= $$\sqrt{1-x^2}$$]

Question 15.
(i) ∫ $$\frac{x-1}{\sqrt{x^2-1}}$$ dx (NCERT)
(ii) ∫ $$\frac{x+2}{\sqrt{x^2-1}}$$ dx (NCERT)
Solution:
(i) Let I = ∫ $$\frac{x-1}{\sqrt{x^2-1}}$$ dx
= $$\int \frac{x d x}{\sqrt{x^2-1}}-\int \frac{d x}{\sqrt{x^2-1}}$$
= I1 – $$\frac{d x}{\sqrt{x^2-1}}$$ …..(1)
where I1 = ∫ $$\frac{x d x}{\sqrt{x^2-1}}$$ ……….(1)
put x2 = t
⇒ 2x dx = dt
∴ I1 = ∫ $$\frac{d t}{2 \sqrt{t-1}}$$
= $$\frac{1}{2}$$ (t – 1)-1/2 dt
= $$\frac{1}{2} \frac{(t-1)^{1 / 2}}{1 / 2}$$
= $$\sqrt{x^{2}-1}$$
I = $$\sqrt{x^2-1}-\log \left|x+\sqrt{x^2-1}\right|$$ + C

(ii) Let I = ∫ $$\frac{x+2}{\sqrt{x^2-1}}$$ dx
= $$\int \frac{x d x}{\sqrt{x^2-1}}+2 \int \frac{d x}{\sqrt{x^2-1}}$$
= $$\frac{1}{2}$$ ∫ (x^{2} – 1)$$-\frac{1}{2}$$ (2x dx) + 2 ∫ $$\frac{d x}{\sqrt{x^2-1^2}}$$
= $$\frac{1}{2} \frac{\left(x^2-1\right)^{\frac{-1}{2}+1}}{\left(\frac{-1}{2}+1\right)}$$ + 2 log |x + $$\sqrt{x^2-1}$$| + C
[∵ ∫ [f(x)]n f'(x) dx = $$\frac{(f(x))^{n+1}}{n+1}$$ + C]
= $$\sqrt{x^2-1}$$ + 2 log |x + $$\sqrt{x^2-1}$$| + C

Question 16.
(i) ∫ $$\sqrt{\frac{a+x}{a-x}}$$ dx (NCERT Exemplar)
(ii) ∫ $$\frac{\sqrt{x^2-a^2}}{x}$$ dx
Solution:
(i) Let I = ∫ $$\sqrt{\frac{a+x}{a-x}}$$ dx
∴ I = ∫ $$\sqrt{\frac{(a+x)}{a-x} \times \frac{a+x}{a+x}}$$ dx
= ∫ $$\frac{a+x}{\sqrt{a^2-x^2}}$$ dx
= ∫ $$a \int \frac{d x}{\sqrt{a^2-x^2}}+\int \frac{x d x}{\sqrt{a^2-x^2}}$$
∴ I = a sin-1 $$\frac{x}{a}$$ + I1
where I1 = ∫ $$\frac{d t}{2 \sqrt{a^2-t}}$$
= $$\frac{1}{2}$$ ∫ (a2 – t)-1/2 dt
= – $$\frac{1}{2} \frac{\left(a^2-t\right)^{1 / 2}}{1 / 2}$$
= – $$\sqrt{a^2-x^2}$$
∴ From (1) ; we have
∴ I = a sin-1 $$\frac{x}{a}$$ – $$\sqrt{a^2-x^2}$$ + C

(ii) Let I =∫ $$\frac{\sqrt{x^2-a^2}}{x}$$ dx

Question 17.
If ∫ $$\frac{2^x}{\sqrt{1-4^x}}$$ dx = k sin-1 (2x) + C, then what is the value of k?
Solution:
Let I = ∫ $$\frac{2^x}{\sqrt{1-4^x}}$$ dx
= ∫ $$\frac{2^x d x}{\sqrt{1-\left(2^x\right)^2}}$$
put 2x = t
⇒ 2x log 2 dx = dt
⇒ I = ∫ $$\frac{d t}{\log 2 \sqrt{1-t^2}}$$
= $$\frac{1}{\log 2} \int \frac{d t}{\sqrt{1-t^2}}$$
I = $$\frac{1}{\log 2}$$ sin-1 t + C
[∵ ∫ $$\frac{d x}{\sqrt{a^2-x^2}}$$ = sin-1 $$\frac{x}{a}$$ + C
∴ I = $$\frac{1}{\log 2}$$ sin-1 (2x) + C ………….(1)
Also given
∫ $$\frac{2^x d x}{1-4 x}$$ = k sin-1 (2x) + C ………..(2)
From (1) and (2) ; we have
k = $$\frac{1}{\log 2}$$