Prove convergence of $\int_0^1 \frac{\ln(x)}{1-x^2} dx$ I'm really lost about how to prove the integral converges.
I tried using the fraction expansion theorem and got to
$$ \frac{\ln(x)}{1-x^2} = \frac{\ln(x)}{2(1+x)} + \frac{\ln(x)}{2(1-x)}$$
But I didn't find a way to use it.
I can't use the Dirichlet's test or Comparing tests since the function is not positive.
What am I missing?
Thank you in advance and have a nice day!
 A: Noticed that $\ln x\leq x-1$ when $0<x\leq 1$,and
$$\int_0^1\frac{x-1}{1-x^2}\mathrm{d}x=\int_0^1\frac{x-1}{(1+x)(1-x)}\mathrm{d}x=-\int_0^1 \frac{\mathrm{d}x}{1+x}=\left[\ln|1+x|\right]_0^1=\ln 2 $$
converges, so $\displaystyle \int_0^1 \frac{\ln x}{1-x^2}\mathrm{d}x$ converges.
A: The intgrand is $f(x):=\dfrac{\ln(x)}{1-x^2}$, with $x\in(0,1)$ so you have
$$f(x)\sim\begin{cases} \ln(x)&&x\in\mathcal U(0^+)\\\dfrac{x-1}{2(1-x)}&&x\in\mathcal U (1^-)\end{cases}$$
and this clearly shows that $\int_{[0,1]}f(x)<+\infty$ since both terms of the asymptotic have convergent integral respectively in a neighbourhoof of $x=0$ and $x=1$.
A: This is definitely not the best approach but it’s quite funnier.
We know that $\sum_{n=1}^{\infty}{1}/{n^2}$ converges (it’s easy to prove), now I will split up this sum into even and odd terms
$$S=\sum_{n=1}^{\infty}\frac{1}{n^2}= \sum_{n=1}^{\infty}\frac{1}{4n^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} $$
$$\implies \sum_{n=1}^{\infty}\frac{1}{n^2} =\frac{4}{3} \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} $$
You can use integration by parts to check that $$\int_1^0 x^{2n}\ln x\,dx=\frac{1}{(2n+1)^2}$$
So our sum becomes
$$S= \frac{4}{3} \sum_{n=0}^{\infty} \int_1^0x^{2n}\ln x\,dx= \frac{4}{3} \int_1^0\ln x \sum_{n=0}^{\infty} x^{2n}\,dx$$
Since $0<x<1$ we can use the geometric series formula, Hence $$S= \frac{4}{3} \int_1^0\frac{\ln x}{1-x^2}\,dx$$
Since $S$ converges, this is immediately implies that $$\int_1^0\frac{\ln x}{1-x^2}\,dx= -\int_0^1\frac{\ln x}{1-x^2}\,dx $$ converges as well.
A: Let us substitute $t = \ln(x)$ that is $x = e^t$ and therefore $\mathrm dx = e^t \mathrm dt$ and thus the integral becomes
$$
\int_{-\infty}^{0}\frac{te^t}{1-e^{2t}}\,\mathrm dt
$$
which can also be written as
$$
-\int_{0}^{\infty}\frac{ze^z}{e^{2z}-1}\,\mathrm dz
$$
by substituting $z = -t$.
Now, when $0<z<\infty$, observe that
$$
\frac{ze^z}{e^{2z}-1} > \frac{ze^z}{e^{2z}}
$$
since denominator is smaller in LHS and therefore
$$
-\frac{ze^z}{e^{2z}-1} < -\frac{ze^z}{e^{2z}}
$$
Now,
$$
\int_{0}^{\infty}ze^{-z}\,\mathrm dz = 1
$$
converges and thus $\displaystyle \int_0^1 \frac{\ln x}{1-x^2}\mathrm{d}x = -\int_{0}^{\infty}\frac{ze^z}{e^{2z}-1}\,\mathrm dz $ also converges.
