Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is wrong. How can I solve this without using the binomial theorem?

Let $u=x^2+1\implies x^2=u-1\implies du=2x\,dx$. So $$\begin{align}\int(x^2+1)^7x^3\,dx&=\int (x^2+1)^7\cdot\frac{x^2}2\cdot2x\,dx\\&=\int u^7\cdot\frac{u-1}2\,du=\frac12\int(u^8-u^7)\,du\\&=\frac12\left(\frac{u^9}9-\frac{u^8}8\right)+C\\&=\boxed{\frac1{144}(x^2+1)^8(8x^2-1)}+C\end{align}$$
