# Evaluate $\rm \int x\tan^{-1}x dx$

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Evaluate $\rm \int x\tan^{-1}x dx$
1. $\rm {1\over2}\left[x^2\tan^{-1}x - x\right]+c$
2. $\rm {1\over2}\left[(x^2+x)\tan^{-1}x - x\right]+c$
3. $\rm {1\over2}\left[(x^2+1)\tan^{-1}x - (x + 1)\right]+c$
4. $\rm {1\over2}\left[(x^2+1)\tan^{-1}x - x\right]+c$
5. None of these

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Correct Answer - Option 4 : $\rm {1\over2}\left[(x^2+1)\tan^{-1}x - x\right]+c$

Concept:

Integration by parts: Integration by parts is a method to find integrals of products.

The formula for integrating by parts is given by:

⇒ $\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times ∫ vdx\right)dx$ + C

where u is the function u(x) and v is the function v(x)

ILATE rule is Usually, the preferred order for this rule and is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

I = $\rm \int x\tan^{-1}x dx$

I = $\rm \tan^{-1}x\int xdx - \int \left({1\over1+x^2}\int xdx\right)dx+c$

I = $\rm {x^2\tan^{-1}x\over2} - {1\over2}\int \left({x^2\over1+x^2}\right)dx+c$

I = $\rm {x^2\tan^{-1}x\over2} - {1\over2}\int \left(1-{1\over1+x^2}\right)dx+c$

I = $\rm {1\over2}\left[x^2\tan^{-1}x - (x-\tan^{-1}x)\right]+c$

I = $\boldsymbol{\rm {1\over2}\left[(x^2+1)\tan^{-1}x - x\right]+c}$