How to solve this definite integral $\ln(1+x^n)$ How to solve this integral: $$\int_{0}^{1} \ln(1+x^n)\,dx.  $$
The problem doesn't say anything about $n$ so I assume $n\in N$.
Source of the question.
 A: Prelude with Harmonic Numbers
$$
\begin{align}
H(x)&=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag1\\
\frac12H\!\left(\frac x2\right)&=\sum_{k=1}^\infty\left(\frac1{2k}-\frac1{2k+x}\right)\tag2\\
H(x)-H\!\left(\frac x2\right)&=\sum_{k=1}^\infty(-1)^{k-1}\left(\frac1k-\frac1{k+x}\right)\tag3
\end{align}
$$
Explanation:
$(1)$: extension of the Harmonic Numbers to $\mathbb{C}$
$(2)$: compute the series for even indices
$(3)$: compute the alternating series
The Harmonic numbers are related to the Digamma function by $H(x)=\gamma+\psi(1+x)$, where $\gamma$ is the Euler-Mascheroni constant.

The Integral
$$
\begin{align}
\int_0^1\log\left(1+x^n\right)\,\mathrm{d}x
&=\sum_{k=1}^\infty\int_0^1\frac{(-1)^{k-1}x^{nk}}k\,\mathrm{d}x\tag4\\
&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k(nk+1)}\tag5\\
&=\sum_{k=1}^\infty(-1)^{k-1}\left(\frac1k-\frac1{k+1/n}\right)\tag6\\[3pt]
&=H\!\left(\frac1n\right)-H\!\left(\frac1{2n}\right)\tag7\\[6pt]
&=\psi\!\left(1+\frac1n\right)-\psi\!\left(1+\frac1{2n}\right)\tag8
\end{align}
$$
Explanation:
$(4)$: apply the Taylor series for $\log(1+x)$
$(5)$: evaluate the integrals
$(6)$: partial fractions
$(7)$: apply $(3)$
$(8)$: give $(7)$ in terms of the Digamma function  
Note that using $(7)$ from this answer, we can compute $(7)$ as a finite sum in terms of logs, sines, and cosines.
A: Integrating by parts as suggested by mathworker21 gives
\begin{align}\int\ln(1+x^n)~\mathrm dx&=x\ln(1+x^n)-n\int\frac{x^n}{1+x^n}~\mathrm dx\\&=x\ln(1+x^n)-n\int\frac{1+x^n-1}{1+x^n}~\mathrm dx\\&=x\ln(1+x^n)-nx+\int\frac n{1+x^n}~\mathrm dx\end{align}
where the last bit is given in this answer as a combination of logarithms and arctangents, provided $n\in\mathbb Z$. In a similar manner, rational $n$ can be handled as well. In the case of irrational $n$, one requires special functions, such as the digamma function.
