The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$  ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$  $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and this is where i got stuck at. I tried to substitute $u =x^7$
but then the integral become $\int \frac{u^2}{28\sqrt{u+1}}du$
the final answer that I found using wolffram calculator is $\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)$

Set $u=x^7$ and $du=7x^6dx$
$$=\frac 1 7 \int u \sqrt{u+1}du$$
Set $\nu=u+1$ and $d\nu=du$
$$\begin{align}
\frac 1 7\int(\nu-1)\sqrt{\nu}d\nu&=\frac 1 7\int\nu ^{3/2}d\nu -\frac 1 7\int\sqrt{\nu}d\nu\\\\
&=\frac{2}{35}(x^7+1)^{5/2}-\frac{2}{21}(x^7+1)^{3/2}+\mathcal C\\\\&=\color{red}{\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)+\mathcal C}\end{align}$$
