Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 32

Given ,
y'= tan^(-1) (x/2)
we have to get the y
so ,
=> y = int (tan^(-1) (x/2)) dx
By Applying the integration by parts we get the solution
so,
let u=tan^(-1) (x/2) => u'= (tan^(-1) (x/2) )'
let t= x/2
=> u' =(du/dt)*(dt/dx) = (d ((tan^(-1)(t))/(dt))(d/dx (x/2))
=(1/(t^2+1))*1/2
=(1/((x/2)^2+1))*1/2
=(4/(x^2+4))*1/2
=2/(x^2+4)

and v'=1 =>v =x
now by Integration by parts ,
int uv' dx= uv-int u'v dx
so , now
int (tan^(-1) (x/2))dx
= xtan^(-1) (x/2) - int 2x/(x^2+4)dx
let x^2+4 = q
=> 2x dx= dq
so ,
int (2x)/(x^2+4)dx = int 1/q dq = ln(q)+c =ln(x^2+4)+c

= x(tan^(-1) (x/2)) -ln(x^2+4)+C