Evaluate $\int_0^\infty \frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\mathrm dx$

Alternatively we can use Feynman's Trick if we let, $$I(a)=\int_0^{\infty} \frac{\ln(ax+1)}{x(x+1)}\mathrm dx,$$ then if we take the derivative of $I(a)$ and expand using partial fractions we have $$I'(a)=\int_0^{\infty} \frac{1}{(ax+1)(x+1)}\mathrm dx =\int_0^{\infty} \frac{a}{(a-1)(ax+1)} - \frac{1}{(a-1)(x+1)}\mathrm dx.$$ Integrating this gives, $$I'(a)=\bigg(\frac{1}{a-1}\ln|ax+1|-\frac{1}{a-1}\ln|x+1| \bigg)|_0^{\infty}=\frac{1}{a-1}\ln|\frac{ax+1}{x+1}| |_0^{\infty}.$$ Noticing the lower bound is zero we're left with an easy limit$$I'(a)=\lim_{x\to\infty}\frac{1}{a-1}\ln|\frac{ax+1}{x+1}|=\frac{\ln(a)}{a-1}.$$ Now if we integrate we can get $I(a)$, this is easy because our integral closely resembles the definition of the dilogarithm, so we have $$I(a)=\int \frac{\ln(a)}{a-1} da=-Li_2(1-a)+C.$$ Now if we notice $I(0)=0$ and note that $Li_2(1)=\frac{\pi^2}{6}$ we see that $C=\frac{\pi^2}{6}.$ So finally if we note that for your problem $a=2$ have that, $$I(2)=-Li_2(-1)+\frac{\pi^2}{6}=\frac{\pi^2}{12}+\frac{\pi^2}{6}=\frac{\pi^2}{4}.$$

This is admittedly not as nice as the approach using the geometric series and makes heavy use of the dilogarithm and it's particular values but I think it's kinda fun.

do the substitution $2x+1\rightarrow x$ and the let $x\rightarrow x^{-1}$ , hence $$\int_{0}^{\infty }\frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\, \mathrm{d}x=2\int_{1}^{\infty }\frac{\ln x}{x^{2}-1}\, \mathrm{d}x=-2\int_{0}^{1}\frac{\ln x}{1-x^{2}}\, \mathrm{d}x$$ then use the geometric series $$\frac{1}{1-x^{2}}=\sum_{n=0}^{\infty }x^{2n}$$ we get $$\int_{0}^{\infty }\frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\, \mathrm{d}x=2\sum_{n=0}^{\infty }\frac{1}{\left ( 2n+1 \right )^{2}}$$ and the answer will follow.