Mathematics

# $\displaystyle\int_{0}^{1}\dfrac{1-x^{2}}{(1+x^{2})^{2}}dx$

##### SOLUTION
Consider, $I=\displaystyle\int_{0}^{1}\dfrac{{\dfrac{1}{x^2}-1}}{\left(x+\dfrac{1}{x}\right)^{2}}\ dx$

Let $x+\dfrac{1}{x}=t$. Then, $d\left(x+\dfrac{1}{x}\right)=dt\Rightarrow \left(1-\dfrac{1}{x^{2}}\right)dx=dt$

Clearly, $x=0\Rightarrow t=\infty$ and $x=1\Rightarrow t=2$

$\therefore I =\displaystyle\int_{\infty}^{2}-\dfrac{1}{t^{2}}dt=\left[\dfrac{1}{t}\right]_{\infty}^{2}=\dfrac{1}{2}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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