$\newcommand{LogI}{\operatorname{Li}}$ We know that the value of $\LogI_{2}(-1)$ is -$\frac{\pi^2}{12}$ and $\LogI_{2}(1)$ is $\frac{\pi^2}{6}$. The value of the polylogarithms has already been addressed here in another question using a method derived by Tom Apostol to prove the Basel Problem: Calculate a $\operatorname{Li}_{2}(-1)$ using Integral Representation.
Using this we can find $$\int_{-1}^{+1}\frac{\ln{(1+t)}}{t}dt=\LogI_{2}(+1)-\LogI_{2}(-1)=\frac{\pi^2}{4}$$
Is there any way we can use Cauchy integral theorem or Residue theorem to find $\int_{-1}^{+1}\frac{\ln{(1+t)}}{t}dt$. The function $\frac{\ln{(1+t)}}{t}$ has a non-isolated singularity at $-1$ and a removable singularity at $0$ which we can subsitute by the $\lim_{z \to 0}\frac{\ln{(1+z)}}{z}=1$.
I think it can be specifially by solving the residue theorem $\lim_{N \to \infty} \int_{\mid z \mid=N+1/2}\frac{z^{-2}}{e^{2i\pi z}-1}dz = 0$ as suggested by @reuns in the one of the comments listed for the question.
