Calculus
posted by Anonymous .
integral (4x^23x+2)/(4x^24x+3) dx

long division shows that we have
∫ 1 + (x1)/(4x^24x+3)
= ∫ 1 + (x1)/[(2x1)^2 + 2] dx
let (2x1)^2 = 2tan^2θ, so
(2x1)^2 + 2 = 2tan^2θ+2 = 2sec^2θ
2x1 = √2 tanθ
x = (√2 tanθ+1)/2
x1 = (√2 tanθ1)/2
dx = 1/2 √2 sec^2θ
∫ 1 + (x1)/[(2x1)^2 + 2] dx
= ∫ (1 + (√2 tanθ1)/(4sec^2θ)) 1/√2 sec^2θ dθ
. . .
= 1/8(x + 1/8 log(4x^24x+3)  1/(4√22) arctan (2x1)/√2