How to integrate $\int x^2(x+1)^9 dx$?

Question

Question: $\int x^2(x+1)^9 dx$

I know it's a simple question but I got different answers by using integration by parts and u-substitution.

u-substitution: let $u=x+1$, then $du=dx$. Then $\int x^2(x+1)^9 dx$=$\int (u-1)^2(u)^9 dx$=$\int u^{11}-2u^{10}+u^9 dx$=$\frac{1}{12}u^{12}-\frac{2}{11}u^{11}+\frac{1}{10}u^{10}+C$, then we substitute $x+1$ in $u$.

Did I do anything wrong?

Answer 1

Use integration by parts twice: $\int x^2(1+x)^9 = (1/10)(1+x)^{10}x^2 -\int (2/10)x(1+x)^{10}$,...

$\int x(1+x)^{10}= (1/11)(1+x)^{11}x-\int (1/11)(1+x)^{11}$

$\int (1+x)^{11}=(1/12)(1+x)^{12}$.