# Is it possible to solve $\int \frac{1}{u^2-u}du$ without hyperbolic tangent?

I'm getting stuck on this integral, and all the tools I see online relate it to hyperbolic tangent. When I try to solve it I break it up using partial fraction decomposition to

$\int \frac{1}{u^2-u}$=$\int \frac{1}{u-1}$ - $\int\frac{1}{u}$ = $$ln(u-1) - ln(u) + C$$ all the solutions online show this integral instead to be $$ln(1-u) -ln(u) +c$$

and I'm trying to reason how either my method was flawed and that solving with hyperbolic tangent is the only way or that ln(u-1) = ln(1-u) which seems impossible to me.

Thanks guys

• It's not the case that $\ln(u-1)=\ln(1-u)$ - but it is true that there's a constant $C$ such that $\ln(u-1)=\ln(1-u)+C$... Feb 4, 2015 at 4:17

You forgot the absolute values in the ln: $\int \frac {1}{x} dx = \ln |x| + C$

• Wow, thanks. That's exactly it. Feb 4, 2015 at 4:13

Your solution is correct if $u\gt 1$. The other version you quote is correct if $1-u\gt 0$ (and $u\gt 0$). You can verify its correctness by differentiating, using the Chain Rule.

Note that if we are working in the reals, $\ln x$ is undefined for $x\le 0$.

Solution $$\int\dfrac{1}{u^2-u}du=2\tanh^{-1}(1-2u)+c$$

Beacuse, use

$$\dfrac{d\tanh^{-1}(x)}{dx}=\dfrac{1}{1-x^2}$$

have

$$\dfrac{d(2\tanh^{-1}(1-2u)+c)}{du}=2\dfrac{-2}{1-(1-2u)^2}=\dfrac{-4}{4u-4u^2}=\dfrac{1}{u^2-u}$$