Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$ 
Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$

So first we have two potentially problematic points which are $1,\infty$
We split the integral to $$\int_1^2 \frac{x\ln x}{x^4-1} dx + \int_2^\infty \frac{x\ln x}{x^4-1} dx$$
Now first I tried to study $I_1 = \int_1^2 \frac{x\ln x}{x^4-1} dx$.
I evaluated $\lim_{x\to 1^+} \frac{x\ln x}{x^4-1} = \frac{1}{4}$ so basically the function is "nice" at $x=1$ and we can evaluate the integral, but how exactly? 
I tried various of methods without luck (integration by parts , substituting $t=\ln x$ , $(x^4 - 1) = (x^2-1)(x^2+1)$ , $(x^4-1) = (x-1)(1+x+x^2+x^3)$) 
EDIT
Following the answer I got, I thought about:
Since $\frac{x\ln(x)}{x^4-1}$ is monotonically decreasing to $0$ then we can use the inequality:
$$\int_1^\infty \frac{x\ln(x)}{x^4-1} \le \sum_{n=1}^\infty \frac{n\ln(n)}{n^4-1} \le \sum_{n=1}^\infty \frac{n^2}{n^4-1} \le \sum_{n=1}^\infty \frac{n^2}{n^4} = \sum_{n=1}^\infty \frac{1}{n^2} \lt \infty$$

For showing, that $I_{2}$ converges, you might use  $\ln(x)\leq x$, which is valid for $x>0$.
