<s>[INST] Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$ While browsing the post Is there any integral for the golden ratio $\phi$?, I came across this nice answer,
$$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi\,\phi}5$$
it seems the general form is just

$$p \int_0^\infty \frac{1}{1+x^{p}}dx=\color{blue}{\frac{\pi}{\sin\big(\tfrac{\pi}{p}\big)}}$$

I wondered about
$$\int_0^\color{red}1 \frac{1}{1+x^p}dx=\,?$$
Mathematica could find messy closed-forms for $p=5,7$. After some laborious simplification, 
$$5\int_0^1 \frac{1}{1+x^5}dx=\frac{\pi\sqrt{\phi}}{5^{1/4}}+\ln2+\sqrt{5}\ln\phi$$
Question 1: In general, is it true that for any $p$ ,

$$2p\,\int_0^1 \frac{1}{1+x^p}dx=\color{blue}{\frac{\pi}{\sin\big(\tfrac{\pi}{p}\big)}}+2\ln2-\psi\big(\tfrac{1}{p}\big)+\psi\big(\tfrac{p-1}{2p}\big)+\psi\big(\tfrac{p+1}{2p}\big)-\psi\big(\tfrac{p-1}{p}\big)$$

where $\psi(z)$ is the digamma function? 
Note: The four digammas, implemented in Mathematica as PolyGamma[z], can be expressed as a sum of cosines x logarithms for odd $p=2m+1$. Let $k=\frac{2n-1}{p}\pi$, then,
$$-\psi\big(\tfrac{1}{p}\big)+\psi\big(\tfrac{p-1}{2p}\big)+\psi\big(\tfrac{p+1}{2p}\big)-\psi\big(\tfrac{p-1}{p}\big)=-4\sum_{n=1}^m \cos (k)\ln\big(\sin\tfrac{k}{2}\big)$$
Question 2: For even $p$, can we can also avoid the digamma by using cosines and logarithms?
 [/INST] I'm only going to address Question 1. 
The expression proposed in Question 1 is true. However, it is a little bit too complicated than necessary. A simpler version of the expression is
$$2p\int_0^1 \frac{dx}{1+x^p} = \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right)$$

From reflection formula, take logarithm and differentiate, we get
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin\pi z} \implies
\psi(z) - \psi(1-z) = \pi\cot\pi z$$
This leads to
$$\frac{\pi}{\sin z} = \pi\cot\frac{\pi z}{2} - \pi\cot\pi z = \psi\left(\frac{z}{2}\right) - \psi\left(1-\frac{z}{2}\right) - \psi(z) + \psi(1-z)
$$
From duplication formula, take logarithm and differentiate, we get
$$\Gamma(z)\Gamma\left(z+\frac12\right) = 2^{1-2z}\sqrt{\pi}\Gamma(2z)
\implies \psi(z) + \psi\left(z + \frac12\right) = -2\log 2 + 2\psi(2z)
$$
Apply these to RHS of Question 1, we can expose RHS to following mess
$$
\left[ \color{red}{\psi\left(\frac{1}{2p}\right)} - \psi\left(1 - \frac{1}{2p}\right) - \color{green}{\psi\left(\frac{1}{p}\right)} + \color{blue}{\psi\left(1 - \frac{1}{p}\right)} \right]
+ \left[ \color{green}{2\psi\left(\frac1p\right)} - \color{red}{\psi\left(\frac{1}{2p}\right)} - \color{magenta}{\psi\left(\frac{p+1}{2p}\right)}\right]\\
- \color{green}{\psi\left(\frac{1}{p}\right)} + \psi\left(\frac{p-1}{2p}\right) + \color{magenta}{\psi\left(\frac{p+1}{2p}\right)} - \color{blue}{\psi\left(\frac{p-1}{p}\right)}
$$
After massive cancellation, we can simplify RHS to
$$
\psi\left(\frac{p-1}{2p}\right) - \psi\left(1 - \frac{1}{2p}\right)
= \psi\left( 1 - \left(\frac{1}{2p} + \frac12\right)\right) - \psi\left(1 - \frac{1}{2p}\right)
= \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right)
$$
Recall following expansion of digamma function
$$\psi(z) = \frac{1}{z} + \sum_{n=1}^\infty \left(\frac{1}{z+n} - \frac{1}{n}\right)$$
We find
$$\begin{align}
\text{RHS}
&= \frac{1}{\frac{1}{2p}} - \frac{1}{\frac{1}{2p} + \frac12} + \sum_{n=1}^\infty\left(\frac{1}{\frac{1}{2p}+n} - \frac{1}{\frac{1}{2p} + n + \frac12}\right)\\
&= 2\sum_{n=0}^\infty\frac{(-1)^n}{\frac{1}{p}+n}
= 2\sum_{n=0}^\infty\int_0^1 (-1)^n t^{\frac{1}{p}+n-1} dt
= 2\int_0^1 \sum_{n=0}^\infty (-1)^n t^{\frac{1}{p}+n-1} dt\\
&= 2 \int_0^1 \frac{t^{\frac{1}{p}-1}}{1+t} dt
= 2p\int_0^1 \frac{dx}{1+x^p} = \text{LHS}
\end{align}
$$
</s>