integrate (ln(2018x+1)/(1+x)^4 dx from 0 to infinte

First, we do integration by parts

$\displaystyle \int \dfrac{\ln(1+2018x)}{(1+x)^{4}} dx = -\dfrac{\ln(1+2018x)}{3(1+x)^{3}}+\dfrac{1}{3} \int \dfrac{2018}{(1+2018x)(1+x)^{4}} dx$

The first term is zero when applying the limit of integration. The second fraction can be written as(using partial fraction technique)

$\displaystyle \int \left ( \dfrac{2018}{3 \cdot 2017} \right ) \left ( \dfrac{2018^{3}}{2017^{2}}\dfrac{1}{1+2018x}- \dfrac{2018^{2}}{2017^{2}}\dfrac{1}{1+x}-\dfrac{2018}{2017}\dfrac{1}{(1+x)^{2}}-\dfrac{1}{(1+x)^{3}}\right )dx$

The exact form is then

$\displaystyle \dfrac{2018}{3 \cdot 2017} \left ( \left ( \dfrac{2018}{2017} \right )^{2} \ln(2018)-\dfrac{2018}{2017}-\dfrac{1}{2} \right ) \approx \boxed{2.039982229}$