How do solve #∫x tan^-1x dx#, given that #d/dx tan^-1x = 1/(1+x^2)# ?

#I=intxtan^-1(x)dx#

We will use integration by parts. Integration by parts takes the form #intudv=uv-intvdu#.

Since we cannot integrate #tan^-1(x)# well, let #u=tan^-1(x)# (notice you've also been provided with the derivative of #tan^-1(x)# in the question). Thus, let #dv# be the remaining portion: #dv=xdx#.

Differentiating #u# and integrating #dv#:

• #u=tan^-1(x)" "=>" "du=1/(1+x^2)dx#
• #dv=xdx" "=>" "v=x^2/2#

Then:

#I=uv-intvdu=x^2/2tan^-1(x)-intx^2/2(1/(1+x^2))dx#

Simplifying slightly:

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

Rewrite the integrand as follows, or perform polynomial long division on #x^2/(1+x^2)#:

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

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

#I=x^2/2tan^-1(x)-1/2intdx-1/2int(-1)/(1+x^2)dx#

Note that #intdx=x#. Furthermore, since #d/dxtan^-1(x)=1/(1+x^2)#, we see that #int1/(1+x^2)dx=tan^-1(x)#.

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

Which we may write as:

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

Rearranging and adding the constant of integration:

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