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

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

Question 1.
(i) \(\int \frac{d x}{x^2+2 x+10}\)
(ii) \(\int \frac{d x}{(x+1)(x+2)}\)
(iii) \(\int \frac{d x}{9 x^2+6 x+1}\)
Solution:
(i) Let I = \(\int \frac{d x}{x^2+2 x+10}\)
= \(\int \frac{d x}{x^2+2 x+1+9}\) = \(\int \frac{d x}{(x+1)^2+3^2}\)
Put x + 1 = t ⇒ dx = dt
= \(\int \frac{d t}{t^2+3^2}\) = \(\frac{1}{3}\)tan^-1\(\left(\frac{t}{3}\right)\)+C
= \(\frac{1}{3}\)tan^-1\(\left(\frac{x+1}{3}\right)\)+C

(ii) Let I = \(\int \frac{d x}{(x+1)(x+2)}\)
= \(\int \frac{d x}{x^2+3 x+2}\) = \(\int \frac{d x}{x^2+3 x+\frac{9}{4}-\frac{1}{4}}\)
= \(\int \frac{d t}{\left(x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}\)
= \(\frac{1}{2 \times \frac{1}{2}} \log \left|\frac{x+\frac{3}{2}-\frac{1}{2}}{x+\frac{3}{2}+\frac{1}{2}}\right|+C\)
= \(\log \left|\frac{x+1}{x+2}\right|+C\)

(iii) Let I = \(\int \frac{d x}{9 x^2+6 x+1}\)
= \(\int \frac{d x}{(3 x+1)^2}\) = \(\int(3 x+1)^{-2} d x\)
= \(\frac{(3 x+1)^{-2+1}}{(-2+1)^3}\) + C = –\(\frac { 1 }{ 3 }\)\(\frac{1}{3 x+1}\)+C

Question 2.
(i) \(\int \frac{d x}{9 x^2+6 x+10}\)
(ii) \(\int \frac{d x}{4 x^2-4 x+3}\)
(iii) \(\int \frac{d x}{1+x-x^2}\)
Solution:
Let I = \(\int \frac{d x}{9 x^2+6 x+10}\)
= \(\frac{1}{9}\)\(\int \frac{d x}{x^2+\frac{2}{3} x+\frac{10}{9}}\)
= \(\frac{1}{9}\)\(\int \frac{d x}{x^2+\frac{2}{3} x+\frac{1}{9}-\frac{1}{9}+\frac{10}{9}}\)
= \(\frac{1}{9}\)\(\int \frac{d x}{x^2+\frac{2}{3} x+\frac{1}{9}-\frac{1}{9}+\frac{10}{9}}\)
= \(\frac{1}{9}\)\(\int \frac{d x}{\left(x+\frac{1}{3}\right)^2+1^2}\)
= \(\frac{1}{9}\) × \(\frac{1}{1}\)tan^-1\(\left(\frac{x+\frac{1}{3}}{1}\right)\) + C
= \(\frac{1}{9}\)tan^-1\(\left(\frac{3 x+1}{3}\right)\) + C

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

(iii) Let I = \(\int \frac{d x}{1+x-x^2}\) = – \(\int \frac{d x}{x^2-x-1}\)
= –\(\int \frac{d x}{x^2-x+\frac{1}{4}-\frac{5}{4}}\)
= –\(\int \frac{d x}{\left(x-\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)}\)
= – \(\frac{1}{2 \times \frac{\sqrt{5}}{2}} \log\)\(\left|\frac{x-\frac{1}{2}-\frac{\sqrt{5}}{2}}{x-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|+C\)
= – \(\frac{1}{\sqrt{5}}\)log\(\left|\frac{2 x-1-\sqrt{5}}{2 x-1+\sqrt{5}}\right|+C\)

Question 3.
\(\int \frac{d x}{x\left(x^6+1\right)}\)
Solution:
Let I = \(\int \frac{d x}{x\left(x^6+1\right)}\) = \(\int \frac{x^5 d x}{x^6\left(x^6+1\right)}\)
Put x⁶ = t ⇒ 6x⁵ dx = dt
= \(\int \frac{d t}{6 t(t+1)}\) = \(\frac{1}{6}\)\(\int \frac{d t}{t^2+t+\frac{1}{4}-\frac{1}{4}}\)
= \(\frac{1}{6}\)\(\int \frac{d t}{\left(t+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\)

Question 4.
\(\int \frac{5 x-2}{1+2 x+3 x^2} d x\)
Solution:
Let I = \(\int \frac{5 x-2}{1+2 x+3 x^2} d x\)
= \(\frac{5}{6}\)\(\int \frac{\left(6 x-\frac{12}{5}\right)}{3 x^2+2 x+1} d x\)
= \(\frac{5}{6}\)\(\int \frac{\left(6 x-\frac{12}{5}\right)}{3 x^2+2 x+1} d x\)
= \(\frac{5}{6}\)\(\int \frac{\left(6 x+2-2-\frac{12}{5}\right)}{3 x^2+2 x+1} d x\)
= \(\frac{5}{6}\)\(\int \frac{6 x+2}{3 x^2+2 x+1} d x\) + \(\frac{5}{6}\)\(\int \frac{-\frac{22}{5}}{3 x^2+2 x+1} d x\)
= \(\frac{5}{6}\)\(\int \frac{d t}{t}\)–\(\frac{11}{3}\)\(\int \frac{d x}{3\left(x^2+\frac{2 x}{3}+\frac{1}{3}\right)}\)
[Put 3x² + 2x + 1 = t ⇒ (6x + 2)dx = dt]
= \(\frac{5}{6}\)log|t| – \(\frac{11}{9}\)\(\int \frac{d x}{\left(x+\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2} d x\)
= \(\frac{5}{6}\)log|3x² + 2x + 11| – \(\frac{11}{9}\) × \(\frac{3}{\sqrt{2}}\)tan^-1\(\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)\) + C
= \(\frac{5}{6}\)log|3x² + 2x + 1| – \(\frac{11}{3 \sqrt{2}}\) tan^-1\(\left(\frac{3 x+1}{\sqrt{2}}\right)\) + C

Question 5.
(i) \(\int \frac{x}{x^4+2 x^2+3} d x\)
(ii) \(\int \frac{e^x}{x^{2 x}+6 e^x+5} d x\)
(iii) \(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x\)
(iv) \(\int \frac{d x}{x\left[10+7 \log x+(\log x)^2\right]}\)
(v) \(\frac{(3 \sin \theta-2) \cos \theta}{5-\cos ^2 \theta-4 \sin \theta} d \theta\)
Solution:
(i) Let I = \(\int \frac{x}{x^4+2 x^2+3} d x\)
Put x² = t ⇒ 2xdx = dt
= \(\int \frac{d t}{2\left(t^2+2 t+3\right)}\) = \(\frac{1}{2}\)\(\int \frac{d t}{t^2+2 t+1+2}\)
= \(\frac{1}{2}\)\(\int \frac{d t}{(t+1)^2+(\sqrt{2})^2}\)
= \(\frac{1}{2}\) × \(\frac{1}{\sqrt{2}}\)tan^-1\(\left(\frac{t+1}{\sqrt{2}}\right)\)
[∵ \(\int \frac{d x}{a^2+x^2}\) = \(\frac{1}{a}\)tan^-1\(\frac{1}{a}\) + C]
= \(\frac{1}{2 \sqrt{2}}\)tan^-1\(\left(\frac{x^2+1}{\sqrt{2}}\right)\) + C

(ii) Let I = \(\int \frac{e^x}{x^{2 x}+6 e^x+5} d x\)
Put e^x = t ⇒ e^xdx = dt
= \(\int \frac{d t}{t^2+6 t+5}\) = \(\int \frac{d t}{t^2+6 t+9-4}\)

(iii) Let I = \(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x\)
Put sin x = t ⇒ cos x dx = dt
= \(\int \frac{d t}{(1-t)(2-t)}\) = \(\int \frac{d t}{t^2-3 t+2}\)
= \(\int \frac{d t}{t^2-3 t+\frac{9}{4}-\frac{9}{4}+2}\)
= \(\int \frac{d t}{\left(t-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}\)
= \(\frac{1}{2 \times \frac{1}{2}}\) log \(\left|\frac{t-\frac{3}{2}-\frac{1}{2}}{t-\frac{3}{2}+\frac{1}{2}}\right|\) + C
[∵ \(\int \frac{d x}{x^2-a^2}\) = \(\frac{1}{2 a}\) log \(\left|\frac{x-a}{x+a}\right|\) + C]
= log \(\left|\frac{t-2}{t-1}\right|\) + C = log \(\left|\frac{\sin x-2}{\sin x-1}\right|\) + C

(iv) Let I = \(\int \frac{d x}{x\left[10+7 \log x+(\log x)^2\right]}\)
Put log x = t ⇒ \(\frac{1}{x}\)dx = dt
= \(\int \frac{d t}{t^2+7 t+10}\)
= \(\int \frac{d t}{t^2+7 t+\frac{49}{4}-\frac{49}{4}+10}\)
= \(\int \frac{d t}{\left(t+\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2}\)
= \(\frac{1}{2 \times \frac{3}{2}}\) log \(\left|\frac{t+\frac{7}{2}-\frac{3}{2}}{t+\frac{7}{2}+\frac{3}{2}}\right|\) + C
= \(\frac{1}{3}\) log \(\left|\frac{t+2}{t+5}\right|\) + C
= \(\frac{1}{3}\) log \(\left|\frac{\log x+2}{\log x+5}\right|\) + C

(v) Let I = \(\frac{(3 \sin \theta-2) \cos \theta}{5-\cos ^2 \theta-4 \sin \theta} d \theta\)
= \(\int \frac{(3 \sin \theta-2) \cos \theta}{5-\left(1-\sin ^2 \theta\right)-4 \sin \theta} d \theta\)
Put sin θ = t ⇒ cos θ dθ = dt
= \(\int \frac{(3 t-2) d t}{4+t^2-4 t}\) = \(\int \frac{(3 t-2)}{(t-2)^2} d t\)
= \(\int \frac{3\left(t-\frac{2}{3}\right)}{(t-2)^2} d t\) = \(\int \frac{\left(t-2+\frac{4}{3}\right)}{(t-2)^2} d t\)
= 3 \(\int \frac{1}{t-2} d t\) + 4\(\int \frac{d t}{(t-2)^2}\)
= 3 log |t – 2| + 4 \(\frac{(t-2)^{-2+1}}{-2+1}\) + C
= 3 log|sin θ – 2| – \(\frac{4}{\sin \theta-2}\) + C