How to Calculate this definite integral or how to solve this series?

I'd like to add a sel-contained answer. We may consider that $$ f(x) = \sum_{n\geq 1}\frac{\sin(nx)}{n} $$ is the $2\pi$-periodic extension of a linear function, which equals $\frac{\pi-x}{2}$ on $(0,2\pi)$. The convergence is uniform over any compact subset of $\mathbb{R}\setminus 2\pi\mathbb{Z}$. If we integrate both sides four times, we get that $$ g(x)=\sum_{n\geq 1}\frac{\sin(nx)}{n^5} $$ is the $2\pi$-periodic extension of a polynomial with degree five, $p(x)=-\frac{x^5}{240}+\frac{\pi x^4}{48}-\frac{\pi ^2 x^3}{36}+\frac{\pi ^4 x}{90}$.
The convergence is uniform over $\mathbb{R}$, hence by evaluating $g$ and $p$ at $\pi/2$ we get $$ \sum_{n\geq 1}\frac{\sin(n\pi/2)}{n^5} = \sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^5} = p\left(\frac{\pi}{2}\right)=\frac{5\pi^5}{1536}.$$ Not by chance, this is related to the value of the wanted integral. By termwise integration

$$ \int_{0}^{1}\log^4(x)\sum_{k\geq 0}(-1)^k x^{2k}\,dx = \sum_{k\geq 0}(-1)^k\int_{0}^{1}x^{2k}\log^4(x)\,dx =\sum_{k\geq 0}\frac{24(-1)^k}{(2k+1)^5}$$ so $$ \int_{0}^{1}\frac{\log^4(x)\,dx}{1+x^2} = 24 p\left(\frac{\pi}{2}\right)=\color{red}{\frac{5\pi^5}{64}}.$$ Summarizing, it is enough to exploit the Fourier series of Bernoulli polynomials.

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As an alternative, we may use Feynman's trick. By Euler's Beta function and the reflection formula for the $\Gamma$ function we have that $$ \int_{0}^{+\infty}\frac{x^s\,dx}{1+x^2} = \frac{\pi}{2}\sec\left(\frac{\pi s}{2}\right)$$ holds for any $s\in(-1,1)$. If we differentiate (with respect to $s$) both sides four times, then perform an evaluation at $s=0$, we get $$ \int_{0}^{1}\frac{\log^4(x)\,dx}{1+x^2}=\frac{3\pi^5}{8}\cdot[z^4]\sec(z)=\frac{3\pi^5}{8}\cdot[z^4]\frac{1}{1-\frac{z^2}{2}+\frac{z^4}{24}}=\frac{3\pi^5}{8}\left(\frac{1}{2^2}-\frac{1}{24}\right) $$ and the conclusion is just the same.

Your approach works perfectly fine! What you have there, in the last line, is the so-called Dirichlet Beta Function $\beta(s)$. What you are looking for in particular is the value of $\beta(5)$ which is in fact expressable in terms of $\pi^5$ yet alone (as linked by Zacky). Similiar to the Riemann Zeta Function, where we have a formula for $\zeta(2n)$, there is a formula for computing $\beta(2n+1)$ for $n\in\Bbb N_0$. In order to be precise we have that

$$\beta(2n+1)~=~(-1)^n\frac{\pi^{2n+1}}{4^{n+1}(2n)!}\operatorname{E}_{2n}\tag1$$

Here $E_n$ denotes a Euler Number. Using $(1)$ you will obtain the value you are looking for. To give some more context. Dr. Sonnhard Graubner gave the value in terms of the Hurwitz Zeta Function $\zeta(s,a)$, which has a quite simple relation to the Dirichlet Beta Function (similiar with the mentioned Lerch Transcendent). Allawonder essentially gave you the integral representation of $\beta(s)$.

Currently I am not aware of a simpler derivation, not relying on $(1)$. However, as with $\beta(3)$ there might exist an elementary way of evaluating the integral or the sum, respectively.