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
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