Need help with $\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx$ I need you help with this integral:
$$\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx.$$
Mathematica says it does not converge, which is apparently false.
Huge pain in the rump, but in the end, fairly straightforward.  First observe that

$$\begin{align}\int_0^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} &amp;= \int_0^{1} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} + \int_1^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)}\\ &amp;= \int_0^{1} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} + \int_0^{1} dx \, x^3 \frac{\log{(1+x)}-\log{x}}{(1+x^2)(1+x^3)}\\ &amp;= \int_0^1 dx \frac{\log{(1+x)}}{1+x^2} - \int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)}\end{align} $$

The first integral may be evaluated by subbing $x=\tan{t}$:

$$\begin{align} \int_0^1 dx \frac{\log{(1+x)}}{1+x^2} &amp;= \int_0^{\pi/4} dt \, \log{(1+\tan{t})}\\ &amp;= \int_0^{\pi/4} dt \, \log{(\sin{t}+\cos{t})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &amp;= \int_0^{\pi/4} dt \, \log{(\sqrt{2} \cos{(t-\pi/4)})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &amp;= \frac{\pi}{8} \log{2} + \int_0^{\pi/4} dt \, \log{(\cos{(t-\pi/4)})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &amp;= \frac{\pi}{8} \log{2}\end{align}$$

The second integral is messy but still doable.  First, we may use partial fractions.  Then we will get a series of zeta-like sums.  Some of them we will immediately recognize.  To the rest, we may apply the residue theorem.

$$\begin{align}\int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)} &amp;= \frac12 \int_0^1 dx \left [\frac{1-x}{1+x^2} - \frac{1-x-x^2}{1+x^3} \right ] \log{x}\\ &amp;= \frac12 \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \left (x^{2 k}-x^{2 k+1}-x^{3 k}+x^{3 k+1}+x^{3 k+2} \right ) \log{x}\\ &amp;= -\frac12 \sum_{k=0}^{\infty} (-1)^k \left [ \frac1{(2 k+1)^2} - \frac1{(2 k+2)^2} - \frac1{(3 k+1)^2}\\ + \frac1{(3 k+2)^2}+\frac1{(3 k+3)^2}\right ] \\ &amp;= \frac{5}{72} \frac{\pi^2}{12} - \frac12 G + \frac12 \sum_{k=0}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] \end{align} $$

where $G$ is Catalan's constant.  As for the sum:

$$\begin{align} \frac12 \sum_{k=0}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] &amp;= \frac14  \sum_{k=-\infty}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] \\ &amp;= -\frac{\pi}{4} \frac1{3^2} \left [\operatorname*{Res}_{z=-1/3} \frac{\csc{\pi z}}{(z+1/3)^2} \\- \operatorname*{Res}_{z=-2/3} \frac{\csc{\pi z}}{(z+2/3)^2} \right ]\\ &amp;= \frac{\pi^2}{36} \left [\frac{\cos{\pi/3}}{\sin^2{\pi/3}} - \frac{\cos{2 \pi/3}}{\sin^2{2 \pi/3}} \right ]\\ &amp;= \frac{\pi^2}{27}\end{align} $$

Therefore

$$\int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)} = \frac{5 \pi^2}{864} - \frac12 G + \frac{\pi^2}{27} = \frac{37 \pi^2}{864} - \frac{G}{2}$$

and finally...

$$\int_0^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} = \frac{G}{2} + \frac{\pi}{8} \log{2} - \frac{37 \pi^2}{864} \approx 0.307524\cdots$$
Hint

This would probably work. Let $u = \ln(1+x)$ then $du = \frac{dx}{1+x}$ and factor the bottom $1+x^3 = (1+x)(\ldots)$.