Mathematics

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


SOLUTION

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

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, 

$$x=0\Rightarrow t=\infty$$ 

$$x=1\Rightarrow t=2$$

$$\therefore =\displaystyle\int_{\infty}^{2}-\dfrac{1}{t^{2}}dt$$ 

$$\left[\dfrac{1}{t}\right]_{\infty}^{2}=\dfrac{1}{2}$$

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