How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried  another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.

Using summation by parts we have $$\sum_{n\leq N}\frac{H_{n}}{n\left(n+1\right)}=H_{N}\sum_{n\leq N}\frac{1}{n\left(n+1\right)}-\sum_{n\leq N-1}\frac{1}{n+1}\left(\sum_{k\leq n}\frac{1}{k\left(k+1\right)}\right).$$ Clearly $$\sum_{n\leq N}\frac{1}{n\left(n+1\right)}=\left(1-\frac{1}{N+1}\right)$$ then $$\sum_{n\leq N}\frac{H_{n}}{n\left(n+1\right)}=H_{N}\left(1-\frac{1}{N+1}\right)-H_{N}+\sum_{n\leq N-1}\frac{1}{\left(n+1\right)^{2}}.$$ Now, since $H_{N}\sim\log\left(N\right)$ as $N\rightarrow+\infty$, the claim follows.
