Evaluate $\int_{0}^{\infty} \frac{\ln(x)}{x^{6}+1} dx$ The hint for this exercise is to use either the gamma or the beta function. I think that a substitution that leads me to the gamma function is probably the way to go because of the bounds of the integral, but I  still haven't found such a substitution. Any hints or suggestions?
 A: Let our integral be $ I$.
To evaluate this integral, define a function $ f : [0,+\infty) \mapsto \mathbb R$
$$ f(a)= \int_0^{\infty} \frac{x^a}{1+x^6}\ \mathrm{d} x$$
Notice that
$$ \left.\frac{\partial}{\partial a }f(a) \right|_{a=0} = I $$
Now, via the substitution $ x^6=t$,
$$ f(a)=\frac{1}{6}\int_0^{\infty} \dfrac{t^{\frac{a+1}{6}-1}}{1+t}\ \mathrm{d} t $$
Via the definition of beta function
$$ B(a,b)=\int_0^{\infty} \frac{x^{a-1}}{(1+x)^{a+b}}\ \mathrm{d}x $$
$$ f(a)=\dfrac{1}{6}B\Bigl(\frac{a+1}{6},1-\frac{a+1}{6}\Bigr) $$
Using the relationship of beta and gamma function
$$ B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
$$ f(a)=\dfrac{\Gamma\Big(\frac{a+1}{6}\Big)\Gamma\Big(1-\frac{a+1}{6}\Big)}{6}$$
Using Euler's reflection formula for Gamma function
$$ \Gamma(z)\Gamma(1-z)=\pi \csc(\pi z)$$
$$f(a)= \dfrac{\pi}{6}\csc\Big(\frac{\pi(a+1)}{6}\Big) $$
Differentiating with respect to $a$,
$$ f^{\prime}(a)= -\frac{\pi ^2}{36}\csc\Big(\frac{\pi(a+1)}{6}\Big)\cot\Big(\frac{\pi(a+1)}{6}\Big) $$
Finally putting $a=0$, we get
$$\boxed{\boxed{I=-\frac{\pi^2}{6\sqrt{3}} }} $$
Please accept the answer if you like it.
