Question #c842e

1 Answer
Feb 1, 2018

#I=1/2(arctan(1/x)x^2+x-arctan(x))+C#

Explanation:

We want to solve

#I=intxarctan(1/x)dx#

Use integration by parts #intudv=uv-intvdu#

Let #u=arctan(1/x)# and #dv=xdx#

Then #du=-1/(x^2+1)dx# and #v=1/2x^2#

#I=1/2arctan(1/x)x^2+1/2intx^2/(x^2+1)dx#

Then solve #I_1=1/2intx^2/(x^2+1)dx#

#I_1=1/2intx^2/(x^2+1)dx#

#=1/2int(x^2+1)/(x^2+1)dx-1/2int1/(x^2+1)dx#

#=1/2x-1/2arctan(x)+C#

Substitute this into the original integral

#I=1/2(arctan(1/x)x^2+x-arctan(x))+C#