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Q.
The value of definite integral $\int\limits_0^1 \frac{1-x}{\left(1+x^2\right)^2} d x$ is equal to
Integrals
Solution:
$\int\limits_0^1 \frac{1-x}{\left(1+x^2\right)^2} d x=\int\limits_0^1 \frac{d x}{(\underset{I_1}{1+x^2})^2}-\int\limits_0^1 \frac{x}{(\underset{I_2}{1+x^2})^2} d x$
$I_1=\int\limits_0^1 \frac{d x}{\left(1+x^2\right)^2} \text { Put } x=\tan \theta, \sec ^2 \theta d \theta=d x $
$\int\limits_0^{\frac{\pi}{4}} \frac{\sec ^2 \theta d \theta}{\sec ^4 \theta}=\int\limits_0^{\frac{\pi}{4}} \cos ^2 \theta d \theta=\frac{1}{2} \int\limits_0^{\frac{\pi}{4}}(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=\int\limits_0^1 \frac{x}{\left(1+x^2\right)^2} d x \text { Put } 1+x^2=t, $
$\therefore 2 x d x=d t, x d x=\frac{d t}{2}$
$=\frac{1}{2} \int\limits_1^2 \frac{ dt }{ t ^2}=\frac{1}{2}\left[-\frac{1}{ t }\right]_1^2=-\frac{1}{2}\left[\frac{1}{2}-1\right]=-\frac{1}{2}\left(-\frac{1}{2}\right)=\frac{1}{4}$
$\therefore I _1- I _2=\frac{\pi}{8}+\frac{1}{4}-\frac{1}{4}=\frac{\pi}{8}$