How do you find the antiderivative of $int (xarctanx) dx$ ?

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How do you find the antiderivative : $int (xarctanx) dx$ ?

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Answer 1 · 2017-09-12T14:28:31

$1/2{(x^2+1)arc tanx-x}+C.$

Suppose that, $I=intx arc tanxdx.$

We will use the following Method of Integration by Parts (IBP) :

IBP : $intu*vdx=uintvdx-int((du)/dxintvdx)dx.$

We take, $u=arc tanx, and, v=x.$

$:. (du)/dx=1/(x^2+1), and, intvdx=x^2/2.$

Therefore, by IBP, we have,

$I=x^2/2*arc tanx-int{x^2/2*1/(x^2+1)}dx,$

$=x^2/2*arc tanx-1/2intx^2/(x^2+1)dx,$

$=x^2/2*arc tanx-1/2int{(x^2+1)-1}/(x^2+1)dx,$

$=x^2/2*arc tanx-1/2int{(x^2+1)/(x^2+1)-1/(x^2+1)}dx,$

$=x^2/2*arc tanx-1/2{int1dx-int1/(x^2+1)dx},$

$=x^2/2*arc tanx-1/2{x-arc tanx}.$

$rArr I=1/2{(x^2+1)arc tanx-x}+C.$

Enjoy Maths.!