Prove by test integrabilty that: $\displaystyle\int_0^{+\infty} \left|\frac{\ln x}{x^2-1}\right|\, dx < +\infty$ how we prove that the function $\dfrac{\ln x}{x^2-1}$ is in $L^1(]0,+\infty[)$ using the tests of convegence integral? I calculate is and i found that $$\displaystyle\int_0^{+\infty} \left|\frac{\ln x}{x^2-1}\right|\, dx &lt; +\infty$$, but i want to prove that this function is in $L^1(]0,+\infty[)$ using the tests of integrability of integral.

Thank you in advance
At infinity use


$$\lim_{x\to\infty} \frac{x^{3/2}\ln x}{x^2-1} = \lim_{x\to\infty} \frac{x^{2}}{x^2-1}\cdot\frac{\ln x}{x^{1/2}} =\lim_{x\to\infty} \frac{1}{2}\cdot\frac{\ln x^{1/2}}{x^{1/2}} = 0$$
That is there exists $A&gt;0$ such that $x&gt;A$ implies 

$$ |\frac{\ln x}{x^2-1}|\le \frac{1}{x^{3/2}}$$


  At $x=0$ use


$$\lim_{x\to0} \frac{x^{1/2}\ln x}{x^2-1} = \frac12 \lim_{x\to0}x^{1/2}\ln x^{1/2} = 0$$
 That is there exists $a&gt;0$ such that $0&lt;x&lt;a$ implies 

$$ |\frac{\ln x}{x^2-1}|\le \frac{1}{x^{1/2}}$$


  At $x=1$ there singularity is fake since, 


$$\lim_{x\to 1} \frac{\ln x}{x-1} = (\ln x)'|_{x=1} =1$$
Notice that $|\frac{\ln x}{x^2-1}| \leq \min\{x^{1/2}, x^{3/2}\}$ and that the RHS is integrable because $p &lt; 1$, $p &gt; 1$ respectively. So by the comparison test, the integral in question is finite, hence $\frac{\ln x}{x^2-1} \in L^1$.