how to integrate (x+1)/(x^2+2x)? the big looking s is in front of that problem?

Answer 1

∫(x+1)/(x^2+2x) dx

∫y'/y dy is ln¦y¦+ c

Notice that x + 1 is almost the derivative of x^2 + 2x.

In fact it is half the derivative.

so ∫(x+1)/(x^2+2x) dx = (1/2) ln¦x^2+2x¦ + c

Answer 2

Write your problem as 1/2*∫ (2x+2) dx / (x²+2x)
Let u = x²+2x, du = (2x+2)dx
So we get 1/2*∫ du/u = 1/2 ln |u|= 1/2 ln|x²+2x| + C.

Answer 3

use it as (X+1) x (X^2+2x)^-1 and integrate it like that