The integral $\ J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx$ Here 
Is there a general formula for $I(m,n)$?
I asked for a general formula for $\ I(m,n):=\int_0^{\infty} \frac{x^m}{x^n+1}dx$.

Is the formula $$I(m,n)=\frac{\pi}{n\sin((m+1)\frac{\pi}{n})}$$ true for every pair $\ (m,n)\ $ of real numbers with $\ 0\le m\le n-2\ $ and not just for non-negative integers with $\ 0\le m\le n-2\ $ ?

I wondered whether there is a similar closed form for
$$J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx$$
I figured out that $$I(m,n)=J(m,n)+J(n-2-m,n)$$ holds for $0\le m\le n-2$.
For $\ m=0\ $, the first few values are
$$J(0,0)=\frac{1}{2}$$ $$J(0,1)=\ln(2)$$ $$J(0,2)=\frac{\pi}{4}$$ $$J(0,3)=\frac{2\sqrt{3}\pi+\ln(64)}{18}$$ $$J(0,4)=\frac{\pi+2\coth^{-1}(\sqrt{2})}{4\sqrt{2}}$$ $$J(0,5)=\frac{1}{5}\sqrt{\frac{1}{10}(5+\sqrt{5})}\pi+\frac{1}{20}(\ln(16)+2\sqrt{5}\coth^{-1}(\frac{3}{\sqrt{5}}))$$ $$J(0,6)=\frac{\pi+\sqrt{3}\ln(2+\sqrt{3})}{6}$$

Is there a general formula for $\ J(m,n)\ $ and does it hold for all pairs $\ (m,n)\ $ of real numbers with $\ 0\le m\le n-2$ ?

 A: There does exist a closed form for
$$
J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx \tag1
$$ in terms of the digamma function $\psi(\cdot)$.

Proposition. Let $m=1,2,\cdots$ and $n=1,2,\cdots$. One has
  $$
J(m,n)=\frac1{2n} \psi\left(\frac{m+n+1}{2n}\right)-\frac1{2n}\psi\left(\frac{m+1}{2n} \right) \tag2
$$ 

then using 
$$
\psi\left(r+1\right)-\psi\left(r \right)=\frac1r,\quad r \in \mathbb{Q}^*,\tag3
$$ and $$
\psi\left(\frac{m}{2n}\right) = -\gamma -\ln(4n) -\frac{\pi}{2}\cot\left(\frac{m\pi}{2n}\right) +2\sum_{k=1}^{n-1} \cos\left(\frac{\pi km}{n} \right) \ln\sin\left(\frac{k\pi}{2n}\right) \quad (m<2n)\tag4
$$ one gets a closed form in terms of a finite number of elementary functions.
Hint. By the change of variable, $x=u^{1/n}$, $dx=\dfrac1n u^{1/n-1}du$, one may write
$$
\begin{align}
J(m,n)&=\int_0^1 \frac{x^m}{x^n+1}dx
\\&=\frac1n \int_0^1 \frac{u^{\frac{m+1}{n}-1}}{1+u}du
\\&=\frac1n \int_0^1 \frac{u^{\frac{m+1}{n}-1}(1-u)}{1-u^2}du
\\&=\frac1{2n} \int_0^1 \frac{v^{\frac{m+n+1}{2n}-1}}{1-v}dv-\frac1{2n} \int_0^1 \frac{v^{\frac{m+1}{2n}-1}}{1-v}dv
\\&=\frac1{2n} \psi\left(\frac{m+n+1}{2n}\right)-\frac1{2n}\psi\left(\frac{m+1}{2n} \right)
\end{align}
$$ then one may conclude with Gauss's digamma theorem.
Edit. For any real numbers $a, b$ such that $a>0$ and $b>0$ we have
$$
\int_0^1 \frac{x^a}{x^b+1}\:dx=\frac1{2b} \psi\left(\frac{a+b+1}{2b}\right)-\frac1{2b}\psi\left(\frac{a+1}{2b} \right) 
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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$\ds{\mrm{J}\pars{m,n} \equiv \int_{0}^{1}{x^{m} \over x^{n} + 1}\,\dd x:\ ?}$.

\begin{align}
\mrm{J}\pars{m,n} & \equiv \int_{0}^{1}{x^{m} \over x^{n} + 1}\,\dd x =
\int_{0}^{1}{x^{m} - x^{m + n} \over 1 - x^{2n}}\,\dd x\quad
\pars{\begin{array}{l}
\mbox{Multiply numerator an denominator}
\\
 \mbox{by}\ds{\quad 1 - x^{n}}
\end{array}}
\\[5mm] &
\stackrel{x^{2n}\ \mapsto\ x}{=}\,\,\,
{1 \over 2n}\int_{0}^{1}{x^{\pars{m + 1}/\pars{2n} - 1} -
x^{\pars{m + n + 1}/\pars{2n} - 1} \over 1 - x}\,\dd x
\\[5mm] & =
\bbox[#ffe,15px,border:1px dotted navy]{\ds{{1 \over 2n}\bracks{%
\Psi\pars{m + n + 1 \over 2n} - \Psi\pars{m + 1 \over 2n}}}}\qquad
\pars{~\Psi:\ Digamma\ Function~}
\end{align}
We used
the identity $\color{#000}{\mathbf{6.3.22}}$: $\ds{\Psi\pars{z} + \gamma =
\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,,\ \Re\pars{z} > 0}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant.
