Evaluate $\int_0^\infty \frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\mathrm dx$ How to evaluate 
$$\int_{0}^{\infty }\frac{\ln(2x+1)}{x(x+1)}\,\mathrm dx?$$
I tried
$$\frac{\ln(2x+1)}{x(x+1)}=\ln(2x+1)\left (\frac{1}{x}-\frac{1}{1+x}  \right)$$
but I don't know how to go on.

This might be a late response, but consider the double integral
$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \frac{x}{1+x^2} \ dy \ dx.$$
We will evaluate $I$ in two ways. Integrating with respect to $y$ first, we see that
$$\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \ dy= \lim_{b \rightarrow \infty}\frac{\arctan((2b+1)x)}{2x}-\frac{\arctan(x)}{2x}=\frac{\pi}{4x}-\frac{\arctan(x)}{2x}$$ and get
$$I=\int_{0}^{\infty} \left(\frac{\pi}{4x}-\frac{\arctan(x)}{2x} \right) \frac{x}{1+x^2} \ dx=\int_{0}^{\infty} \frac{\pi}{4(1+x^2)}-\frac{\arctan(x)}{2(1+x^2)} \ dx=\frac{\pi^2}{8}-\frac{\pi^2}{16}=\frac{\pi^2}{16}.$$ 
Now we reverse the order of integration as such:
$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+x^2(2y+1)^2} \frac{x}{1+x^2} \ dx \ dy.$$
To clean things up a little bit, let $u=x^2, \ du = 2 x \ dx.$ So 
$$I=\int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{1+u(2y+1)^2} \frac{u}{2(1+u)} \ du \ dy.$$
Integrating with respect to $u,$ we can use partial fractions to get that 
$$\frac{1}{1+u(2y+1)^2} \frac{1}{2(1+u)}=\frac{(2y+1)^2}{2(1+u(2y+1)^2)((2y+1)^2-1)}-\frac{1}{2(u+1)((2y+1)^2-1)}.$$
We see 
$$\int_{0}^{\infty} \frac{(2y+1)^2}{2(1+u(2y+1)^2)((2y+1)^2-1)}-\frac{1}{2(u+1)((2y+1)^2-1)} \ du $$
$$= \lim_{b \rightarrow \infty}\frac{\ln(b(2y+1)^2+1)}{2(2y+1)^2-2} - \frac{\ln(b+1)}{2(2y+1)^2-2}= \frac{\ln(2y+1)}{(2y+1)^2-1}=\frac{\ln(2y+1)}{4y(y+1)}.$$
Hence, 
$$I=\int_{0}^{\infty}\frac{\ln(2y+1)}{4y(y+1)} \ dy=\frac{\pi^2}{16},$$
So $$I=\int_{0}^{\infty}\frac{\ln(2y+1)}{y(y+1)} \ dy=\frac{\pi^2}{4}.$$
