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?
 A: 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 $$\int (x^2 + 1)^7 x^3 \, dx = \frac{1}{2} \int u^7 (u-1) \, du = \frac{1}{2}  \int u^8 - u^7 \, du = \frac{1}{2} \left(\frac{u^9}{9} - \frac{u^8}{8}\right) + C = \frac{(x^2+1)^8}{18} \left( x^2 - \frac{1}{8} \right) + C.$$
A: Simply set $y = x^2+1$, then $\frac{dy}{dx} = 2x$ and therefore
$$\int (x^2+1)^7x^3 dx = \int y^7\frac{1}{2}(y-1)dy$$
I think the rest is trivial.
A: 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. 
A: Let $u=x^2+1\implies x^2=u-1\implies du=2x\,dx$. So $$\begin{align}\int(x^2+1)^7x^3\,dx&=\int (x^2+1)^7\cdot\frac{x^2}2\cdot2x\,dx\\&=\int u^7\cdot\frac{u-1}2\,du=\frac12\int(u^8-u^7)\,du\\&=\frac12\left(\frac{u^9}9-\frac{u^8}8\right)+C\\&=\boxed{\frac1{144}(x^2+1)^8(8x^2-1)}+C\end{align}$$
A: by part method
$$\int(x^2 + 1)^7x^3dx=x^2*\frac{(x^2+1)^8}{2}-\int(x^2 + 1)^8xdx$$
$$x^2*\frac{(x^2+1)^8}{2}-\frac{1}{18}(x^2+1)^9+C$$
A: By definition of integral without substitutions:
$$(x^2+1)^7x^3=(x^2+1)^7(x^2+1-1)x=(x^2+1)^8x-(x^2+1)^7x.$$
