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

Question

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

2508 views around the world

You can reuse this answer
Creative Commons License

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.!