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 $\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)$
 A: Hint: Try $u=x^7+1$. $\,\,\,\,\,$
A: Set $u=x^7$ and $du=7x^6dx$
$$=\frac 1 7 \int u \sqrt{u+1}du$$
Set $\nu=u+1$ and $d\nu=du$
$$\begin{align}
\frac 1 7\int(\nu-1)\sqrt{\nu}d\nu&=\frac 1 7\int\nu ^{3/2}d\nu -\frac 1 7\int\sqrt{\nu}d\nu\\\\
&=\frac{2}{35}(x^7+1)^{5/2}-\frac{2}{21}(x^7+1)^{3/2}+\mathcal C\\\\&=\color{red}{\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)+\mathcal C}\end{align}$$
A: Hint: Use the substitution $u^2=x^7+1$
A: \begin{align}
\int x^{13}\sqrt{x^7+1}\,dx&=\int x^6(x^7+1)\sqrt{x^7+1}\,dx-\int x^6\sqrt{x^7+1}\,dx\\
&=\int x^6(x^7+1)^{3/2}\,dx-\int x^6(x^7+1)^{1/2}\,dx
\end{align}
Then, use $u=x^7+1$ in order to evaluate these integrals.
A: The integrand is a special case of the more general form $$f(x;m,n,a) = x^{2n-1} (x^n + a)^{1/m}.$$  We observe that the choice $$u^m = x^n + a, \quad m u^{m-1} \, du = n x ^{n-1} \, dx$$ along with $x^{2n-1} = \frac{1}{n} x^n (nx^{n-1})$ gives $$\int f(x;m,n,a) \, dx = \frac{1}{n} \int (u^m - a) u \cdot mu^{m-1} \, du = \frac{m}{n} \left(\frac{u^{2m+1}}{2m+1} - \frac{au^{m+1}}{m+1} \right) + C,$$ which after substituting back, gives $$\frac{m}{n}(x^n+a)^{1 + 1/m} \left( \frac{x^n+a}{2m+1} - \frac{a}{m+1} \right) + C.$$  The given integral is the special case $m = 2$, $n = 7$, $a = 1$.
