How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$ $$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$
How to find sum of this series?
I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that  I am not able to solve this.

$$\sum_{n=1}^\infty\frac1{n(2n+1)}=\sum_{n=1}^\infty\frac1n\int_0^1 x^{2n}dx=\int_0^1\sum_{n=1}^\infty\frac{x^{2n}}{n}dx=-\int_0^1\ln(1-x^2)dx$$
$$=-\underbrace{\int_0^1\ln(1-x)dx}_{1-x=y}-\underbrace{\int_0^1\ln(1+x)dx}_{1+x=y}=-\int_0^2\ln(y)dy$$
$$=y-y\ln(y)|_0^2=2-2\ln(2)$$
