Question
Evaluate the following integrals:
$\int^\limits1_0\frac{1-\text{x}^2}{(1+\text{x}^2)^2}\text{ dx}$

Answer

Let $\text{I}=\int^\limits1_0\frac{1-\text{x}^2}{(1+\text{x}^2)^2}\text{ dx}$ Then,
$\text{I}=\int^\limits1_0\frac{\big(\frac{1}{\text{x}^2}-1\big)}{\big(\text{x}+\frac{1}{\text{x}}\big)^2}\text{ dx}$
Let $\text{x}+\frac{1}{\text{x}}=\text{t}$ Then, $1-\frac{1}{\text{x}^2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=\infty$ and $\text{x}=1,\text{t}=2$
$\therefore\ \text{I}=\int^\limits2_\infty\frac{-\text{dt}}{\text{t}^2}$
 $\Rightarrow\text{I}=\Big[\frac{1}{\text{t}}\Big]^2_\infty$
$\Rightarrow\text{I}=\frac{1}{2}-0$
$\Rightarrow\text{I}=\frac{1}{2}$

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