We need to find the integral of f(x) = 2x/(x+1)*(x^2+1)

f(x) = 2x/(x+1)*(x^2+1)

finding the partial fractions

=> f(x) = (Ax + B)/(x^2 + 1) + C/(x + 1)

=> f(x) = [Ax^2 + Ax + Bx + B + Cx^2 + C]/(x^2 +1)(x +1)

=> f(x) = [x^2(A + C) + x(A + B) + B + C]/(x^2 +1)(x +1)

=> A + C = 0 , A + B = 2 and B + C = 0

A = -C

B - C = 2 and B + C = 0

=> 2B = 2

=> B = 1

C = -1

A = 1

This gives f(x) = 2*x/(x^2+1) - 1/(x+1) + 1/(x^2+1)

Int[f(x) dx]

=> Int[2*x/(x^2+1) - 1/(x+1) + 1/(x^2+1)]

=> 2*Int [x/(x^2 + 1) dx] - Int[1/(x + 1) dx] + Int[1/(x^2 + 1) dx]

=> 2*Int [x/(x^2 + 1) dx] - ln(x + 1) + arc tan x

Int[x/(x^2 + 1) dx]

let x^2 + 1 = y

dy/2 = x dx

=> (1/2)*Int [ 1/y dy]

=> (1/2)*ln y

=> (1/2)*ln ( x^2 + 1)

The required integral is (1/2)*ln (x^2 + 1) - ln(x + 1) +arc tan x