Question
Evaluate the following integrals:$\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
✓
Answer
Let $\text{I}=\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$ Let $\text{x}^2=\text{t}$ Then, $2\text{xdx}=\text{dt}$ When $\text{x}=10,\text{t}=0$ and $\text{x}=1,\text{t}=1$$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{1}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=\tan^{-1}1-\tan^{-1}0$
$\Rightarrow\text{I}=\frac{\pi}{4}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{1}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=\tan^{-1}1-\tan^{-1}0$
$\Rightarrow\text{I}=\frac{\pi}{4}$
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