Question

The value of definite integral $\underset{0}{\overset{1}{\int }}\frac{1-x}{{\left(1+x^2\right)}^2}dx$ is equal to

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{12}$

Answer 1

Answer: (A)

$\underset{0}{\overset{1}{\int }}\frac{1-x}{{\left(1+x^2\right)}^2}dx=\underset{0}{\overset{1}{\int }}\frac{dx}{(\underset{I_1}{1+x^2}{)}^2}-\underset{0}{\overset{1}{\int }}\frac{x}{(\underset{I_2}{1+x^2}{)}^2}dx$
$I_1=\underset{0}{\overset{1}{\int }}\frac{dx}{{\left(1+x^2\right)}^2}\text{ Put }x=\tan \theta ,{\sec }^2\theta d\theta =dx$
$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{{\sec }^2\theta d\theta }{{\sec }^4\theta }=\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\cos }^2\theta d\theta =\frac{1}{2}\underset{0}{\overset{\frac{\pi }{4}}{\int }}(1+\cos 2\theta )d\theta =\frac{1}{2}{\left[\theta +\frac{\sin 2\theta }{2}\right]}_0^{\frac{\pi }{4}}=\frac{1}{2}\left[\frac{\pi }{4}+\frac{1}{2}\right]=\frac{\pi }{8}+\frac{1}{4}$
$I_2=\underset{0}{\overset{1}{\int }}\frac{x}{{\left(1+x^2\right)}^2}dx\text{ Put }1+x^2=t,$
$\therefore 2xdx=dt,xdx=\frac{dt}{2}$
$=\frac{1}{2}\underset{1}{\overset{2}{\int }}\frac{dt}{t^2}=\frac{1}{2}{\left[-\frac{1}{t}\right]}_1^2=-\frac{1}{2}\left[\frac{1}{2}-1\right]=-\frac{1}{2}\left(-\frac{1}{2}\right)=\frac{1}{4}$
$\therefore I_1-I_2=\frac{\pi }{8}+\frac{1}{4}-\frac{1}{4}=\frac{\pi }{8}$