Question

Integral dari x^2 + 2x / (x+1)^2 dx

Answer 1

Integral dari (x² + 2x)dx = Integral (x²+2x)(x+1)^-2dx
                     (x + 1)²
Misal U = x² + 2x ⇒ du = (2x + 2)dx ⇒ (1/2)du = (x + 1)dx
Integral (x²+2x)dx = Integral U du
                                           (1/2)²
                           = Integral U/(1/4)du
                           = Integral 4Udu
                           = 2U² + C
                           = 2(x² + 2x)² + C

Answer 2

Perhatikan bahwa :
(x^2+2x)/(x+1)^2=((x+1)^2-1)/(x+1)^2 = (x+1)^2/(x+1)^2 - 1/(x+1)^2
= 1 - 1/(x+1)^2
so,
= int (1 - 1/(x+1)^2) dx
= int dx - int 1/(x+1)^2 dx
= x - int 1/(x+1)^2 dx

misal :
u = x + 1
du = dx
so,
= x - int 1/u^2 du
= x  - int u^-2  du
= x + u^-1 + C
= x + 1/u + C
= x + 1/(x+1)  + C