Question

The value of definite integral
(see attachment)​

Answer 1

Given to find,

$\small\text{ \displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{x\tan^{-1}x}{1+e^{\tan^{-1}x}}\ dx\quad\dots(1) }$

We know the following property of definite integral,

• $\small\displaystyle\int\limits_a^bf(x)\ dx=\int\limits_a^bf(a+b-x)\ dx$

On applying this property we get,

$\small\text{ \displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{(-1+1-x)\tan^{-1}(-1+1-x)}{1+e^{\tan^{-1}(-1+1-x)}}\ dx }$

$\small\text{ \displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{-x\tan^{-1}(-x)}{1+e^{\tan^{-1}(-x)}}\ dx }$

$\small\text{ \displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{-x(-\tan^{-1}x)}{1+e^{-\tan^{-1}x}}\ dx }$

$\small\text{ \displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{x\tan^{-1}x}{1+e^{-\tan^{-1}x}}\ dx }$

$\small\text{ \displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{x\tan^{-1}x}{\left(1+\dfrac{1}{e^{\tan^{-1}x}}\right)}\ dx }$

$\small\text{ \displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{x\tan^{-1}x}{\left(\dfrac{e^{\tan^{-1}x}+1}{e^{\tan^{-1}x}}\right)}\ dx }$

$\small\text{ \displaystyle\longrightarrow I=\int\limits_{-1}^1\dfrac{e^{\tan^{-1}x}\cdot x\tan^{-1}x}{1+e^{\tan^{-1}x}}\ dx\quad\dots(2) }$

Adding (1) and (2),

$\small\text{ \displaystyle\longrightarrow I+I=\int\limits_{-1}^1\dfrac{x\tan^{-1}x}{1+e^{\tan^{-1}x}}\ dx+\int\limits_{-1}^1\dfrac{e^{\tan^{-1}x}\cdot x\tan^{-1}x}{1+e^{\tan^{-1}x}}\ dx }$

$\small\text{ \displaystyle\longrightarrow 2I=\int\limits_{-1}^1\left(\dfrac{x\tan^{-1}x}{1+e^{\tan^{-1}x}}+\dfrac{e^{\tan^{-1}x}\cdot x\tan^{-1}x}{1+e^{\tan^{-1}x}}\right)\ dx }$

$\small\text{ \displaystyle\longrightarrow 2I=\int\limits_{-1}^1\dfrac{\left(1+e^{\tan^{-1}x}\right)x\tan^{-1}x}{1+e^{\tan^{-1}x}}\ dx }$

$\small\displaystyle\longrightarrow 2I=\int\limits_{-1}^1x\tan^{-1}x\ dx$

We need to perform integration by parts according to ILATE rule where,

• I - Inverse

• L - Logarithmic

• A - Algebraic

• T - Trigonometric

• E - Exponential

Here $\small\tan^{-1}x$ is an Inverse function and $\smallx$ is an Algebraic function, so integration by parts is done by treating $\small\tan^{-1}x$ first, then $\smallx,$ like,

$\small\displaystyle\longrightarrow 2I=\left[\tan^{-1}x\int x\ dx\right]_{-1}^1-\int\limits_{-1}^1\dfrac{1}{1+x^2}\left(\int x\ dx\right)\ dx$

$\small\displaystyle\longrightarrow 2I=\left[\tan^{-1}x\cdot\dfrac{x^2}{2}\right]_{-1}^1-\int\limits_{-1}^1\dfrac{1}{1+x^2}\cdot\dfrac{x^2}{2}\ dx$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[x^2\tan^{-1}x\right]_{-1}^1-\dfrac{1}{2}\int\limits_{-1}^1\dfrac{x^2}{1+x^2}\ dx$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[x^2\tan^{-1}x\right]_{-1}^1-\dfrac{1}{2}\int\limits_{-1}^1\dfrac{1+x^2-1}{1+x^2}\ dx$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[x^2\tan^{-1}x\right]_{-1}^1-\dfrac{1}{2}\int\limits_{-1}^1\left(1-\dfrac{1}{1+x^2}\right)\ dx$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[x^2\tan^{-1}x\right]_{-1}^1-\dfrac{1}{2}\left[x-\tan^{-1}x\right]_{-1}^1$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[x^2\tan^{-1}x-x+\tan^{-1}x\right]_{-1}^1$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[(x^2+1)\tan^{-1}x-x\right]_{-1}^1$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[(1^2+1)\tan^{-1}1-1\right]-\dfrac{1}{2}\left[((-1)^2+1)\tan^{-1}(-1)-(-1)\right]$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[2\left(\dfrac{\pi}{4}\right)-1\right]-\dfrac{1}{2}\left[2\left(-\dfrac{\pi}{4}\right)+1\right]$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[\dfrac{\pi}{2}-1\right]-\dfrac{1}{2}\left[-\dfrac{\pi}{2}+1\right]$

$\small\displaystyle\longrightarrow 2I=\dfrac{1}{2}\left[\dfrac{\pi}{2}-1\right]+\dfrac{1}{2}\left[\dfrac{\pi}{2}-1\right]$

$\small\displaystyle\longrightarrow 2I=\dfrac{\pi}{2}-1$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{I=\dfrac{\pi}{4}-\dfrac{1}{2}}} }$