How do solve #∫x tan^-1x dx#, given that #d/dx tan^-1x = 1/(1+x^2)# ?
1 Answer
Explanation:
#I=intxtan^-1(x)dx#
We will use integration by parts. Integration by parts takes the form
Since we cannot integrate
Differentiating
#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
#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
#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#