The integral $\ J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx$

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 $$ \int_0^1 \frac{x^a}{x^b+1}\:dx=\frac1{2b} \psi\left(\frac{a+b+1}{2b}\right)-\frac1{2b}\psi\left(\frac{a+1}{2b} \right) $$

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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.