Question

The value of the integral $\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}x}{1+x^2}dx$ is _______________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}x}{1+x^2}dx \\ \mathrm{Let}, \\ {\tan }^{-1}x=t \\ \Rightarrow d\left({\tan }^{-1}x\right)=dt \\ \Rightarrow \frac{1}{1+x^2}dx=dt \\ \mathrm{Also},\mathrm{if}x=0,t=0 \\ \mathrm{if}x=1,t=\frac{\mathrm{\pi }}{4} \\ \mathrm{Thus}, \\ I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}tdt \\ ={\left[\frac{t^2}{2}\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ =\frac{1}{2}\left[{\left(\frac{\mathrm{\pi }}{4}\right)}^2-{\left(0\right)}^2\right] \\ =\frac{1}{2}\left[\frac{{\mathrm{\pi }}^2}{16}\right] \\ =\frac{{\mathrm{\pi }}^2}{32} \end{gathered}$$

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Hence, the value of the integral $\underset{0}{\overset{1}{\int }}\frac{{\tan }^{-1}x}{1+x^2}dx\mathrm{is}\underline{\frac{{\mathrm{\pi }}^2}{32}}.$