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.
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
= lnx  lnx + 1 + C
= ln(x)/(x + 1) + C
∫dx/(x)(x + 1)
∫[(x + 1  x)]/(x)(x + 1) dx
∫(1/x  1/(x + 1)) dx
= lnx  lnx + 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.
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.