Integral of $\frac{1}{t^{2}-1}$

Question

When I solve the indefinite integral $\int \frac{1}{t^{2}-1}$ with partial fraction decomposition I obtain $$\int \frac{1}{t^{2}-1} dt=\int \frac{1}{2(t-1)} - \frac{1}{2(t+1)} dt=\frac{1}{2}(\ln(t-1)-\ln(t+1))+C$$

But when I compare it with Wolfram Alpha, they have $$\frac{1}{2}(\ln(1-t)-\ln(t+1))+C$$

$\ln(t-1) \neq \ln(1-t)$ , did I calculate something wrong or?

Answer 1

They're both wrong. It should be$$\frac12\left(\ln(|t-1|-\ln|t+1|\right)+C.$$