<s>[INST] 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?
 [/INST] Quite simply, $$\int (x^2 + 1)^7 x^3 \, dx = \frac{1}{2} \int (x^2 + 1)^7 (x^2)(2x) \, dx,$$ and with the substitution $u = x^2 + 1$, $du = 2x \, dx$, we readily obtain $$\int (x^2 + 1)^7 x^3 \, dx = \frac{1}{2} \int u^7 (u-1) \, du = \frac{1}{2}  \int u^8 - u^7 \, du = \frac{1}{2} \left(\frac{u^9}{9} - \frac{u^8}{8}\right) + C = \frac{(x^2+1)^8}{18} \left( x^2 - \frac{1}{8} \right) + C.$$
</s>