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.

Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.
Noting that
$$\int_0^1 x^{n - 1} \, dx = \frac{1}{n} \quad \text{and} \quad \int_0^1 y^n \, dy = \frac{1}{n + 1},$$
the sum can be written as
$$\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} = \int_0^1 \int_0^1 \frac{1}{x} \sum_{n = 1}^\infty H_n (xy)^n \, dx dy.\tag1$$
Here the order of the summation with the double integral has been changed and can be done due to the dominated convergence theorem. 
Making use of the generating function for the Harmonic number, namely
$$-\frac{\log(1-x)}{1 - x}~=~\sum_{n=1}^\infty H_n x^n,$$
the infinity sum appearing inside the integral in (1) can be written as
$$\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} = -\int_0^1 \int_0^1 \frac{\ln (1 - xy)}{x(1 - xy)} \, dx dy.$$
Enforcing a substitution of $x \mapsto x/y$ gives
\begin{align}
\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} &= -\int_0^1 \int_0^y \frac{\ln (1 - x)}{x (1 - x)} \, dx dy\\
&= -\int_0^1 \int_0^y \left [\frac{\ln (1 - x)}{x} + \frac{\ln (1 - x)}{1 - x} \right ] \, dx \, dy\\
&= \int_0^1 \operatorname{Li}_2 (y) \, dy + \frac{1}{2} \int_0^1 \ln^2 (1 - y) \, dy\\
&= I_1 + \frac{1}{2} I_2.
\end{align}
Here $\operatorname{Li}_2 (y)$ is the dilogarithm function.
For the first of these integrals, writing the dilogarithm function in terms of its series representation, we have
\begin{align}
I_1 &= \int_0^1 \sum_{n = 1}^\infty \frac{y^n}{n^2} \, dy\\
&= \sum_{n = 1}^\infty \frac{1}{n^2} \int_0^1 y^n \, dy\\
&= \sum_{n = 1}^\infty \frac{1}{n^2(n + 1)}\\
&= \sum_{n = 1}^\infty \left [\frac{1}{n^2} + \frac{1}{n + 1} - \frac{1}{n} \right ]\\
&= \sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \left (\frac{1}{n} - \frac{1}{n + 1} \right )\\
&= \frac{\pi^2}{6} - 1,
\end{align}
where the first of the series is the well-known Basel problem while the second telescopes. 
For the second of the integrals, define
$$I(a) = \int_0^1 (1 - y)^a \, dy, \quad a > - 1.$$
Observe that
$$I''(0) = \int_0^1 \ln^2 ( 1 - y) \, dy,$$
where the derivative is with respect to $a$. Now, as
$$I(a) = \operatorname{B} (1, a + 1) = \frac{\Gamma (a + 1)}{\Gamma (a + 2)} = \frac{1}{a + 1},$$
where $\operatorname{B}(x,y)$ is the Beta function, one readily has
$$I''(a) = \frac{2}{(a + 1)^3},$$
yielding $I''(0) = I_2 = 2$. 
Thus
$$\sum_{n = 1}^\infty \frac{H_n}{n (n + 1)} = \frac{\pi^2}{6} - 1 + \frac{1}{2} \cdot 2 = \frac{\pi^2}{6}.$$
