Question

How do this solve ∫ x+an^-1 x dx ?


Upper bound (to): 2
∫
Lower bound (from): 0

Answer 1

Given to find,

$\small\displaystyle\longrightarrow I=\int\limits_0^2x\tan^{-1}x\ dx$

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,

$\small\displaystyle\longrightarrow I=\left[\tan^{-1}x\int x\ dx\right]_0^2-\int\limits_0^2\dfrac{1}{1+x^2}\left(\int x\ dx\right)\ dx$

$\small\displaystyle\longrightarrow I=\left[\tan^{-1}x\cdot\dfrac{x^2}{2}\right]_0^2-\int\limits_0^2\dfrac{1}{1+x^2}\cdot\dfrac{x^2}{2}\ dx$

$\small\displaystyle\longrightarrow I=\dfrac{1}{2}\left[x^2\tan^{-1}x\right]_0^2-\dfrac{1}{2}\int\limits_0^2\dfrac{x^2}{1+x^2}\ dx$

$\small\displaystyle\longrightarrow I=\dfrac{1}{2}\left[x^2\tan^{-1}x\right]_0^2-\dfrac{1}{2}\int\limits_0^2\dfrac{1+x^2-1}{1+x^2}\ dx$

$\small\displaystyle\longrightarrow I=\dfrac{1}{2}\left[x^2\tan^{-1}x\right]_0^2-\dfrac{1}{2}\int\limits_0^2\left(1-\dfrac{1}{1+x^2}\right)\ dx$

$\small\displaystyle\longrightarrow I=\dfrac{1}{2}\left[x^2\tan^{-1}x\right]_0^2-\dfrac{1}{2}\left[x-\tan^{-1}x\right]_0^2$

$\small\displaystyle\longrightarrow I=\dfrac{1}{2}\left[x^2\tan^{-1}x-x+\tan^{-1}x\right]_0^2$

$\small\displaystyle\longrightarrow I=\dfrac{1}{2}\left[(x^2+1)\tan^{-1}x-x\right]_0^2$

$\small\displaystyle\longrightarrow I=\dfrac{1}{2}\left[(2^2+1)\tan^{-1}2-2\right]-\dfrac{1}{2}\left[(0^2+1)\tan^{-1}0-0\right]$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{I=\dfrac{5}{2}\tan^{-1}2-1}} }$

Answer 2

→ Hey Mate:-

→ Given Question:-

→ How do this solve ∫ x+an^-1 x dx ?

→ Upper bound (to): 2

→ ∫

→ Lower bound (from): 0

→ Solution:-

→ Kindly Refer to the attachment , done by me.