How do I take the integral of 1/[x^2+x] with respect to x

I've tried integration by parts, but I keep ending up with 0=0 at the end. I know the answer is ln(x)-ln(x+1)+C, but I need to know how to reach this.
I saw that wolfram used a hyperbolic tangent function to do this, but is there another way? We haven't had a lecture over hyperbolic functions yet.

∫dx/(x² + x)

∫dx/(x)(x + 1)

∫[(x + 1 - x)]/(x)(x + 1) dx

∫(1/x - 1/(x + 1)) dx

= ln|x| - ln|x + 1| + C

= ln|(x)/(x + 1)| + C

You don't have to use hyperbolic trig functions. MechEng has the right idea. What he is using (it isn't obvious from his solution) is partial fraction decomposition.

Note

1/(x² + x) = 1/[x(x + 1)].

You have a proper rational function where the denominator is the product of two linear factors. Put

1/(x² + x) = A/x + B/(x + 1).

Multiply through by x² + x to get

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

Then A = 1 and A + B = 0 ==> B = -1.

Now integrate

∫ 1/(x² + x) dx = ∫ 1/x dx - ∫ 1/(x + 1) dx.

It should be straight forward from here.

FYI: If you have a rational function in the integrand, partial fractions may be the way to go. It doesn't always work, but it should come to mind.