what is the integral of 1/(x^2+x)?

Answer 1

1 / (x) (x + 1) = A / x + B / (x + 1)
1 = A(x + 1) + Bx
1 = (A + B) x + A

A + B = 0
A = 1

A = 1 , B = - 1

I = ∫(1/x) dx - ∫ 1 / (x + 1) dx
I = log x - lg (x + 1) + log K
I = log ( K x / (x + 1 )

Answer 2

1/(x²+x)
= 1/(x(x+1))
= A/x + B/(x+1)

A(x+1) + Bx = 1
If x=0:
A = 1

If x= -1
B = -1

1/(x²+x)
= 1/x - 1/(x+1)

∫1/(x²+x) dx
= ∫ 1/x - 1/(x+1) dx
= ln(x) - ln(x+1) + c

Answer 3

log(x)-log(x +1)