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?

By definition of integral without substitutions:
$$(x^2+1)^7x^3=(x^2+1)^7(x^2+1-1)x=(x^2+1)^8x-(x^2+1)^7x.$$
