Divergent looking but....

$\int_0^∞ x^3 \log(1+e^x) -x^4 dx = \int_0^∞ x^3 (\log(1+e^x)-x) dx$ $= \int_0^∞ x^3 (\log(\frac{1+e^x}{e^x})) dx = \int_0^∞ x^3 \log(1+e^{-x}) dx$ $= \sum_{n=1}^∞ (-1)^{n+1} \frac 1n \int_0^∞ x^3 e^{-nx} dx$ Take $nx=u$ , the Integral becomes $\sum_{n=1}^∞ (-1)^{n+1} \frac{1}{n^5} \int_0^∞ u^3 e^{-u} du = \sum_{n=1}^∞ (-1)^{n+1} \frac{1}{n^5} \Gamma{(4)}$ $=3!( \frac{1}{1^5} -\frac{ 1} {2^5} +\frac {1 }{3^5} -\cdots)$ $=\color{#E81990} \boxed{3!.\frac{15}{16} \zeta{(5)} = \frac{45}{8} \zeta{(5)}}$

$\boldsymbol{{Note}}\::$ $\mathrm{1}\boldsymbol{+}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\boldsymbol{+}\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\boldsymbol{+}\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{5}} }\boldsymbol{+}...\boldsymbol{=}\zeta\left(\mathrm{5}\right)$ $\mathrm{1}\boldsymbol{-}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\boldsymbol{+}\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\boldsymbol{-}\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{5}} }\boldsymbol{+}..\boldsymbol{=}{S}\:\:\:$ And $S+\zeta{(5)} = 2(\frac{1}{1^5} + \frac{1}{3^5} +\frac{1}{5^5} +...) = 2\sum_{n=0}^∞\frac{1}{(2n+1)^5}$ Generally , $\sum_{n=0}^∞ \frac{1}{(2n+1)^5} +\sum_{n=1}^∞ \frac{1}{(2n)^5} = \sum_{n=1}^∞\frac{1}{n^5}$ So, $2\sum_{n=0}^∞ \frac{1}{(2n+1)^{5}} =2 \zeta{(5)}(1-\frac{1}{2^5}) = S+\zeta{(5)}$ $S= \zeta{(5)} (1-\frac{1}{2^4}) = \boxed{\frac{15}{16}\zeta{(5)}}$ Answer is $\color{#20A900}45(8-5) =135$