Question

${\int }_0^{1}x{\tan }^{-1}\left(x\right)dx=$

• A. $\frac{\pi }{4}$
• B. $\left(\frac{\pi }{4}+\frac{1}{2}\right)$
• C. $\left(\frac{\pi }{4}-\frac{1}{2}\right)$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$\left(\frac{\pi }{4}-\frac{1}{2}\right)$

Explanation for correct option :

Let $I={\int }_0^{1}x\tan \left(x\right)dx$

Using the ILATE rule (integration by parts) we will get ${\tan }^{-1}\left(x\right)$ as first function and $x$ as second function, therefore,

$\int uvdx=u\int vdx-\int \left[\frac{du}{dx}\int vdx\right]dx$

Hence using the above formula for the integration, we get

$$\begin{gathered} I={\tan }^{-1}\left(x\right){\int }_0^{1}xdx-{\int }_0^1\left[\frac{d{\tan }^{-1}\left(x\right)}{dx}{\int }_0^{1}xdx\right]dx \\ ={\left[\frac{x^2}{2}{\tan }^{-1}\left(x\right)\right]}_0^1-{\int }_0^1\left[\frac{d{\tan }^{-1}\left(x\right)}{dx}{\int }_0^{1}xdx\right]dx\left[\because \int x^{n}dx=\frac{x^{n+1}}{n+1}\right] \\ =\left[\frac{1}{2}{\tan }^{-1}\left(1\right)-0\right]-{\int }_0^1\left[\frac{1}{1+x^2}\times \frac{x^2}{2}\right]dx\left[\because \frac{d{\tan }^{-1}\left(x\right)}{dx}=\frac{1}{1+x^2}\right] \\ =\left[\frac{1}{2}{\tan }^{-1}\left(\tan \left(\frac{\pi }{4}\right)\right)\right]-\frac{1}{2}{\int }_0^1\left[\frac{x^2+1-1}{1+x^2}\right]dx \\ =\left[\frac{1}{2}\times \frac{\pi }{4}\right]-\frac{1}{2}{\int }_0^1\frac{1+x^2}{1+x^2}dx+\frac{1}{2}{\int }_0^1\frac{1}{1+x^2}dx \\ =\frac{\pi }{8}-\frac{1}{2}{\left[x\right]}_0^1+\frac{1}{2}{\left[{\tan }^{-1}\left(x\right)\right]}_0^1 \\ =\frac{\pi }{8}-\frac{1}{2}+\frac{1}{2}\left[{\tan }^{-1}\left(1\right)-0\right] \\ =\frac{\pi }{8}-\frac{1}{2}+\frac{1}{2}\left[{\tan }^{-1}\left(\tan \left(\frac{\pi }{4}\right)\right)\right] \\ =\frac{\pi }{8}-\frac{1}{2}+\frac{1}{2}\left[\frac{\pi }{4}\right] \\ =\frac{\pi }{8}+\frac{\pi }{8}-\frac{1}{2} \\ I=\frac{\pi }{4}-\frac{1}{2} \end{gathered}$$

Therefore the value of the given integral is equal to $\frac{\pi }{4}-\frac{1}{2}$

Hence, option (C) is the correct answer.