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?

You have chosen the wrong functions for the integration by parts. Take: $$f'(x)=x(x^2+1)^7\\ f(x)=\frac{1}{16}(x^2+1)^8$$
and $$g(x)=x^2\\ g'(x)=2x$$ Hence:
\begin{align}
\int x^3(x^2+1)^7\,dx = \frac{1}{16}(x^2+1)^8 x^2-\int \frac{x}{8}(x^2+1)^8\,dx
\end{align}
And the last one is easy to do. 
