Question
Evaluate the following integrals:
$\int\limits_{0}^{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$
$\int\limits_{0}^{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$
✓
Answer
Let $\text{I}=\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$ Then,
Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{4}}\text{t}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{\text{t}^2}{2}\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{\pi^2}{32}$
Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{4}}\text{t}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{\text{t}^2}{2}\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{\pi^2}{32}$
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