OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(a)

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S Chand Class 12 ICSE Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(a)

Question 1.
(i) \(\int \frac{1}{x^2+36} d x\)
(ii) \(\int \frac{d x}{1+\frac{x^2}{4}}\)
(iii) \(\int \frac{d x}{50+2 x^2}\)
Solution:
(i) \(\int \frac{1}{x^2+36} d x\) = \(\int \frac{1}{x^2+6^2} d x\)
= \(\frac{1}{6}\)tan^-1 \(\frac{x}{6}\) + C
[∵ \(\int \frac{d x}{x^2+a^2}\) = \(\frac{1}{a}\)tan^-1 \(\frac{x}{a}\) + C]

(ii) \(\int \frac{d x}{1+\frac{x^2}{4}}\) = \(\int \frac{4}{2^2+x^2} d x\)
= \(\frac{4}{2}\)tan^-1\(\frac{x}{2}\) = 2 tan^-1\(\frac{x}{2}\) + C

(iii) \(\int \frac{d x}{50+2 x^2}\) = \(\frac{1}{2}\)\(\int \frac{d x}{x^2+5^2}\)
= \(\frac{1}{2}\) × \(\frac{1}{5}\) tan^-1\(\left(\frac{x}{5}\right)\) + C
= \(\frac{1}{10}\) tan^-1 \(\left(\frac{x}{5}\right)\) + C

Question 2.
(i) \(\int \frac{d x}{x^2-4}\)
(ii) \(\int \frac{d x}{9 x^2-16}\)
(iii) \(\int \frac{d y}{25-16 y^2}\)
(iv) \(\int \frac{d y}{18-2 x^2}\)
Solution:

Question 3.
(i) \(\int \frac{d x}{(x+2)^2+1}\)
(ii) \(\int \frac{d x}{1+2(x+2)^2}\)
Solution:
(i) Let I = \(\int \frac{d x}{(x+2)^2+1}\);
Put x + 2 = t ⇒ dx = dt
= \(\int \frac{d t}{t^2+1^2}\) = tan^-1\(\left(\frac{t}{1}\right)\) + C
= tan^-1(x + 1) + C

(ii) Let I = \(\int \frac{d x}{1+2(x+2)^2}\);
Put x + 2 = t ⇒ dx = dt

Question 4.
(i) \(\int \frac{3 x^2}{x^6+1} d x\)
(ii) \(\int \frac{x^2}{1+x^4} d x\)
(iii) \(\int \frac{\cos x}{1+\sin ^2 x} d x\)
(iv) \(\int \frac{e^x}{1+e^{2 x}} d x\)
(v) \(\int \frac{d x}{e^x+e^{-x}}\)
(vi) \(\int \frac{e^{-x}}{16+9 e^{-2 x}} d x\)
(vii) \(\int \sqrt{e^x-1} d x\)
Solution:
(i) Let I = \(\int \frac{3 x^2 d x}{x^6+1}\) = \(\int \frac{3 x^2 d x}{\left(x^3\right)^2+1}\)
Put x³ = t ⇒ 3x²dx = dt
∴ I = \(\int \frac{d t}{t^2+1^2}\) = \(\frac { 1 }{ 2 }\)tan^-1t + C
= tan^-1 x³ + C

(ii) Let I = \(\int \frac{x d x}{1+x^4}\);
Put x² = t ⇒ 2xdx = dt
= \(\int \frac{d t}{2\left(1+t^2\right)}\) = \(\frac { 1 }{ 2 }\)tan^-1t + C
= \(\frac { 1 }{ 2 }\)tan^-1(x²) + C

(iii) Let I = \(\int \frac{\cos x d x}{1+\sin ^2 x}\);
Put sin x = t ⇒ cos x dx = dt
= \(\int \frac{d t}{1+t^2}\) = tan^-1t+C
= tan^-1(sin x)+C

(iv) Let I = \(\int \frac{e^x}{1+e^{2 x}} d x\);
Put e^x = t ⇒ e^xdx = dt
∴ I = \(\int \frac{d t}{1+t^2}\) = \(\frac { 1 }{ 1 }\)tan^-1t+C
= tan^-1(e^x)+C

(v) Let I = \(\int \frac{d x}{e^x+e^{-x}}\) = \(\int \frac{e^x d x}{e^{2 x}+1}\);
Put e^x = t ⇒ e^xdx = dt
= \(\int \frac{d t}{t^2+1}\) = tan^-1t+C
= tan^-1(e^x)+C

(vi) Let I = \(\int \frac{e^{-x}}{16+9 e^{-2 x}} d x\);
Put e^x = t ⇒ e^xdx = dt
= \(\int \frac{-d t}{16+9 t^2}\) = –\(\frac { 1 }{ 9 }\)\(\int \frac{d t}{t^2+(4 / 3)^2}\)
= –\(\frac { 1 }{ 9 }\)×\(\frac{1}{4 / 3}\)tan^-1\(\left(\frac{t}{4 / 3}\right)\)+C
= –\(\frac { 1 }{ 12 }\)tan^-1\(\left(\frac{3 e^{-x}}{4}\right)\)+C

(vii) Let I = \(\int \sqrt{e^x-1} d x\);
Put \(\sqrt{e^x-1}\) = t ⇒ e^x-1 = t²
⇒ e^x = t² + 1 ⇒ e^xdx = 2t dt
⇒ dx = \(\frac{2 t d t}{t^2+1}\)
∴ I = \(\int \frac{2 t^2 d t}{t^2+1}\)
= \(2 \int\left[1-\frac{1}{t^2+1}\right] d t\)
= \(2\left[t-\tan ^{-1} t\right]+C\)
= \(\left[\sqrt{e^x-1}-\tan ^{-1} \sqrt{e^x-1}\right]+C\)

Question 5.
(i) \(\int \frac{x^2-1}{x^2+4} d x\)
(ii) \(\int \frac{x^4}{x^2+1} d x\)
(iii) \(\int \frac{3 x^5}{1+x^{12}} d x\)
(iv) \(\int \frac{d x}{2+\cos x} d x\)
Solution:

(ii) Let I = \(\int \frac{x^4}{x^2+1} d x\)
= \(\int \frac{x^4-1+1}{x^2+1} d x\)
= \(\int \frac{x^4-1}{x^2+1}\)dx + \(\int \frac{d x}{x^2+1}\)
= \(\int\left(x^2-1\right) d x\) + \(\int \frac{d x}{x^2+1^2}\)
= \(\frac{x^3}{3}\) – x + tan^-1x + C

(iii) Let I = \(\int \frac{3 x^5}{1+x^{12}} d x\);
Put x⁶ = t ⇒ 6x⁵dx = dt
= \(\frac { 3 }{ 6 }\)\(\int \frac{d x}{1+t^2}\) = \(\frac { 1 }{ 2 }\)tan^-1t+C
= \(\frac { 1 }{ 2 }\)tan^-1x⁶+C

(iv) Let I = \(\int \frac{d x}{2+\cos x}\)
Put tan \(\frac { x }{ 2 }\) = t ⇒ sec²\(\frac { x }{ 2 }\)\(\frac { 1 }{ 2 }\)dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)

Question 6.
(i) \(\int \frac{x d x}{x^4-a^4}\)
(ii) \(\int \frac{x^2 d x}{x^6-a^6}\)
(iii) \(\int \frac{x}{1-x^4} d x\)
Solution:
(i) Let I = \(\int \frac{x d x}{x^4-a^4}\);
Put x² = t ⇒ 2x dx = dt

(ii) Let I = \(\int \frac{x^2 d x}{a^6-x^6}\);
Put x³ = t ⇒ 3x²dx = dt

(iii) Let I = \(\int \frac{x}{1-x^4} d x\);
Put x² = t ⇒ 2x dx = dt
= \(\frac { 1 }{ 2 }\)\(\int \frac{d t}{1-t^2}\) = \(\frac { 1 }{ 2 }\)\(\int \frac{d t}{1^2-t^2}\)

Question 7.
(i) \(\int \frac{x^3+x}{x^4-9} d x\)
(ii) \(\int \frac{\cos x}{4-\sin ^2 9} d x\)
(iii) \(\int \log \left(2+x^2\right) d x\)
(iv) \(\int \frac{x^2-4}{x^4+16} d x\)
Solution:
(i) Let I = \(\int \frac{x^3+x}{x^4-9} d x\) = \(\int \frac{x\left(x^2+1\right)}{x^4-9} d x\)
= \(\int \frac{x^3}{x^4-9} d x+\int \frac{x d x}{x^4-9}\)
= \(\frac{1}{4} \int \frac{4 x^3 d x}{x^4-9}+\int \frac{d t}{2\left(t^2-9\right)}\) [Put x² = t ⇒ 2x dx = dt]

(ii) Let I = \(\int \frac{\cos x}{4-\sin ^2 x} d x\);
Put sin x = t ⇒ cos x dx = dt
= \(\int \frac{d t}{4-t^2}\) = \(\int \frac{d t}{2^2-t^2}\)

(iii) Let I = \(\int \log \left(2+x^2\right) \cdot 1 d x\);
= log(2 + x²) x – \(\int \frac{2 x}{2+x^2} x d x\)
= x log(2 + x²) – 2\(\int \frac{2+x^2-2}{2+x^2} d x\)
= x log(2+x²) – 2\(\int\left[1-\frac{2}{2+x^2}\right] d x\)
= x log(2 + x²) – 2x + 4\(\int \frac{d x}{x^2+(\sqrt{2})^2}\)
= x log(2 + x²) – 2x + 4\(\int \frac{d x}{x^2+(\sqrt{2})^2}\)
= x log(2 + x²) – 2x + \(\frac{4}{\sqrt{2}}\)tan^-1\(\left(\frac{x}{\sqrt{2}}\right)\)+C

(iv) Let I = \(\int \frac{x^2-4}{x^4+16} d x\);
Divide numerator & deno. by x²; we have

Question 8.
(i) \(\int \frac{x^2}{x^4+1} d x\)
(ii) \(\int \frac{d x}{x^4+1}\)
(iii) \(\int \frac{d x}{x^4+16}\)
(iv) \(\int \frac{x^2}{x^4+16} d x\)
(v) \(\int \frac{\left(x^2+1\right) d x}{x^4+7 x^2+1}\)
(vi) \(\int \frac{x^2+4}{x^4+16} d x\)
Solution:
(i) Let I = \(\int \frac{x^2}{x^4+1} d x\) = \(\frac { 1 }{ 2 }\)\(\int \frac{2 x^2}{x^4+1} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{\left(x^2+1\right)+\left(x^2-1\right)}{x^4+1}\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{x^2+1}{x^4+1} d x\) + \(\frac { 1 }{ 2 }\)\(\int \frac{x^2-1}{x^4+1} d x\)
= \(\frac { 1 }{ 2 }\)I₁ + \(\frac { 1 }{ 2 }\)I₂ …(1)
where I₁ = \(\int \frac{x^2+1}{x^4+1} d x\);
divide Numerator & deno by x²; we have

(ii) Let I = \(\int \frac{d x}{x^4+1}\) = \(\frac { 1 }{ 2 }\)\(\int \frac{2 d x}{x^4+1}\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{\left(x^2+1\right)-\left(x^2-1\right)}{x^4+1} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{x^2+1}{x^4+1} d x\) – \(\frac { 1 }{ 2 }\)\(\int \frac{x^2-1}{x^4+1} d x\)
= \(\frac { 1 }{ 2 }\)I₁ + \(\frac { 1 }{ 2 }\)I₂ …(1)
Where I₁ = \(\int \frac{x^2+1}{x^4+1} d x\);
divide numerator & deno by x²; we have

(iii) Let I = \(\int \frac{d x}{x^4+16}\) = \(\frac { 1 }{ 8 }\)\(\int \frac{8 d x}{x^4+16}\)
= \(\frac { 1 }{ 8 }\)\(\int \frac{\left(x^2+4\right)-\left(x^2-4\right)}{x^4+16} d x\)
= \(\frac { 1 }{ 8 }\)\(\int \frac{x^2+4}{x^4+16} d x\) – \(\frac { 1 }{ 2 }\)\(\int \frac{x^2-4}{x^4+16} d x\)
= \(\frac { 1 }{ 8 }\)I₁ – \(\frac { 1 }{ 8 }\)I₂ ….(1)
Where I₁ = \(\int \frac{x^2+4}{x^4+16} d x\);
Divide Numerator & deno by x², we have

Divide numerator and demo. by x²; we have

put x + \(\frac { 4 }{ x }\) = u ⇒ \(\left(1-\frac{4}{x^2}\right)\) dx = du
on squaring both sides, we have
x² + \(\frac{16}{x^2}\) + 8 = u²
⇒ x² + \(\frac{16}{x^2}\) = u² – 8

(iv) Let I = \(\int \frac{x^2}{x^4+16} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{2 x^2}{x^4+16} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{\left(x^2+4\right)+\left(x^2-4\right)}{x^4+16} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{x^2+4}{x^4+16} d x\) + \(\frac { 1 }{ 2 }\)\(\int \frac{x^2-4}{x^4+16} d x\)
= \(\frac { 1 }{ 2 }\)I₁ + \(\frac { 1 }{ 2 }\)I₂
Let I = \(\int \frac{d x}{x^4+1}\) = \(\frac { 1 }{ 2 }\)\(\int \frac{2 d x}{x^4+1}\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{\left(x^2+1\right)-\left(x^2-1\right)}{x^4+1} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{x^2+1}{x^4+1} d x\) – \(\frac { 1 }{ 2 }\)\(\int \frac{x^2-1}{x^4+1} d x\)
= \(\frac { 1 }{ 2 }\)I₁ – \(\frac { 1 }{ 2 }\)I₂ …(1)
Where I₁ = \(\int \frac{x^2+1}{x^4+1} d x\);
divide Numerator & deno by x²; we have

(v) Let I = \(\int \frac{\left(x^2+1\right) d x}{x^4+7 x^2+1}\)
Divide numerator and deno. by x²; we have

(vi) Let I = \(\int \frac{x^2+4}{x^4+16} d x\)
Divide numerator and deno. by x²; we have

Question 9.
(i) \(\int \frac{x^2+1}{x^4+x^2+1} d x\)
(ii) \(\int \frac{x^2+9}{x^4-2 x^2+81} d x\)
(iii) \(\int \frac{x^2-8}{x^4+7 x^2+64} d x\)
Solution:
(i) Let I = \(\int \frac{x^2+1}{x^4+x^2+1} d x\)
Divide numerator and deno by x²; we have

Question 10.
(i) \(\int \frac{x^2}{x^4+x^2+1} d x\)
(ii) \(\int \frac{d x}{x^4+x^2+1}\)
Solution:
(i) Let I = \(\int \frac{x^2}{x^4+x^2+1} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{2 x^2}{x^4+x^2+1} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{\left(x^2+1\right)+\left(x^2-1\right)}{x^4+x^2+1} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{x^2+1}{x^4+x^2+1} d x\) + \(\frac { 1 }{ 2 }\)\(\int \frac{x^2-1}{x^4+x^2+1} d x\)
= \(\frac { 1 }{ 2 }\)I₁ – \(\frac { 1 }{ 2 }\)I₂ …(1)
Where I₁ = \(\int \frac{x^2+1}{x^4+x^2+1} d x \text {; }\)
Divide numerator and deno. by x²

Put x – \(\frac { 1 }{ x }\) = t ⇒ \(\left(1+\frac{1}{x^2}\right)\)dx = dt
on squaring both sides; we have
\(\left(x-\frac{1}{x}\right)^2\) = t² ⇒ x² + \(\frac{1}{x^2}\) – 2 = t²
⇒ x² + \(\frac{1}{x^2}\) = t² + 2

Divide numerator and deno by x²; we have

(ii) Let I = \(\int \frac{d x}{x^4+x^2+1}\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{2 d x}{x^4+x^2+1} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \frac{\left(x^2+1\right)-\left(x^2-1\right)}{x^4+x^2+1} d x\)
= \(\frac { 1 }{ 2 }\)I₁ – \(\frac { 1 }{ 2 }\)I₂
= \(\frac { 1 }{ 2 }\) \(\int \frac{x^2+1}{x^4+x^2+1} d x\) + \(\frac { 1 }{ 2 }\)\(\int \frac{x^2-1}{x^4+x^2+1} d x\)
where I₁ = \(\int \frac{x^2+1}{x^4+x^2+1} d x\);
Divide numerator and demo. by x²; we have

Question 11.
\(\int \frac{(x-1)^2}{x^4+x^2+1} d x\)
Solution:
Let I = \(\int \frac{(x-1)^2}{x^4+x^2+1} d x\) = \(\int \frac{x^2-2 x+1}{x^4+x^2+1} d x\)
= \(\int \frac{x^2+1}{x^4+x^2+1} d x\) = \(\int \frac{2 x}{x^4+x^2+1} d x\)
= I₁ – I₂ …(1)
∴ I₁ = \(\int \frac{x^2+1}{x^4+x^2+1} d x\);
divide Num & Demo by x²; we have

Put x – \(\frac { 1 }{ x }\) = t ⇒ \(\left(1+\frac{1}{x^2}\right)\)dx = dt
on squaring both sides; we have
\(\left(x-\frac{1}{x}\right)^2\) = t²
⇒ x² + \(\frac{1}{x^2}\) – 2 = t²
⇒ x² + \(\frac{1}{x^2}\) = t² + 2

Putting eqn. (2) & eqn. (3) in eqn.(1); we have
I = \(\frac{1}{\sqrt{3}}\)tan^-1\(\left(\frac{x^2-1}{x \sqrt{3}}\right)\) – \(\frac{2}{\sqrt{3}}\)tan^-1\(\left(\frac{2 x^2+1}{\sqrt{3}}\right)\)+C
where C₁ – C₁ = C