Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 27

Given ,
int tan^(-1) (x)dx
By Applying the integration by parts we get this solution
so,
let u=tan^(-1) (x) =>
u'= (tan^(-1) (x) )'=1/(x^2+1)

and v'=1 =>v =x
now by Integration by parts ,
int uv' dx= uv-int u'v dx
so , now
int tan^(-1) (x)dx
= xtan^(-1) (x) - int 1/(x^2+1)*x dx
= xtan^(-1) (x) - int x/(x^2+1)dx
= xtan^(-1) (x) - 1/2int 2x/(x^2+1)dx
let q=x^2+1
=> dq = 2x dx
so ,
1/2int 2x/(x^2+1)dx = 1/2int 1/q dq = 1/2ln(q)+C
=1/2ln(x^2+1)+c
so, now
int tan^(-1) (x)dx= xtan^(-1) (x) -1/2ln(x^2+1)+C