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?

The following close-form
holds for any integer $p\ge 2$
\begin{align}
\int_0^1 \frac1{1+x^p}dx
= \frac2p \sum_{k=1}^{[\frac p2]} ( \theta_k \sin2\theta_k + \cos2\theta_k \ln \cos\theta_k)
\end{align}
where $\theta_k= \frac{p-2k+1}{2p}\pi $. In particular
\begin{align}
 & \int_0^1 \frac1{1+x^5} dx= \frac{\pi\sqrt{\phi}}{5^{5/4}}+\frac15\ln2+\frac1{\sqrt5}\ln\phi\\
 & \int_0^1 \frac1{1+x^6} dx= \frac\pi6 +\frac1{2\sqrt3}\ln(2+\sqrt3) \\
 & \int_0^1 \frac1{1+x^7} dx 
=\frac\pi{14}\csc \frac{\pi}7-\frac27\left(\frac{\ln\cos\frac{3\pi}7}{\sec\frac{\pi}7} +\frac{\ln\cos\frac{2\pi}7}{\sec\frac{3\pi}7} -\frac{\ln\cos\frac{\pi}7}{\sec\frac{2\pi}7} \right)
\end{align}
