evaluate the integral $\int_{0}^{+\infty} \frac{\log x}{x^4+1}\mathrm{d}x$ I have no idea how to even go about this one. I tried to set $z^4+1 = 0$ and I don't even know if my answers are correct. Someone help please!!!
 A: Using $\gamma=[0,R]\cup Re^{i\frac\pi2[0,1]}\cup i[R,0]$ as $R\to\infty$, we get
$$
\begin{align}
\int_\gamma\frac{\log(z)}{z^4+1}\,\mathrm{d}z
&=\int_0^\infty\frac{\log(x)}{x^4+1}\,\mathrm{d}x-i\int_0^\infty\frac{\frac\pi2i+\log(x)}{x^4+1}\,\mathrm{d}x\\
&=\int_0^\infty\frac{\log(x)}{x^4+1}\,\mathrm{d}x+\frac\pi2\int_0^\infty\frac1{x^4+1}\,\mathrm{d}x-i\int_0^\infty\frac{\log(x)}{x^4+1}\,\mathrm{d}x
\end{align}
$$
There is one singularity at $e^{i\pi/4}$ inside $\gamma$:
$$
\operatorname*{Res}_{z=e^{i\pi/4}}\frac{\log(z)}{z^4+1}
=\frac{i\pi/4}{4e^{i3\pi/4}}
=\frac\pi{16\sqrt2}(1-i)
$$
Therefore,
$$
\begin{align}
\int_0^\infty\frac{\log(x)}{x^4+1}\,\mathrm{d}x+\frac\pi2\int_0^\infty\frac1{x^4+1}\,\mathrm{d}x-i\int_0^\infty\frac{\log(x)}{x^4+1}\,\mathrm{d}x
&=2\pi i\frac\pi{16\sqrt2}(1-i)\\
&=\frac{\pi^2}{8\sqrt2}(1+i)
\end{align}
$$
Which gives us not only
$$
\int_0^\infty\frac{\log(x)}{x^4+1}\,\mathrm{d}x=-\frac{\pi^2}{8\sqrt2}
$$
but also
$$
\int_0^\infty\frac1{x^4+1}\,\mathrm{d}x=\frac\pi{2\sqrt2}
$$
A: 
I have no idea how to even go about this one.

Notice that if $~I(a)=\displaystyle\int_0^\infty\frac{x^{^{a-1}}}{x^n+1}dx,~$ then $~\displaystyle\int_0^\infty\frac{\ln x}{x^n+1}dx=I'(1).~$ But $I(a)$ can be shown
to equal $\dfrac\pi n\cdot\csc\bigg(a\cdot\dfrac\pi n\bigg)$ by first letting $t=\dfrac1{x^n+1}$ , then recognizing the expression of the beta
function in the new integral, and lastly applying Euler's reflection formula for the $\Gamma$ function to the new result. Now all that's left to do is to differentiate the expression with regards to a, then replace
a with $1$ and n with $4$. The final result should be $-\bigg(\dfrac\pi n\bigg)^2\cdot\csc\bigg(a\cdot\dfrac\pi n\bigg)\cdot\cot\bigg(a\cdot\dfrac\pi n\bigg)$, which in our case becomes $-\dfrac{\pi^2}{8\sqrt2}$ .
