Use integration by substitution I'm trying to evaluate integrals using substitution. I had  
$$\int (x+1)(3x+1)^9 dx$$
My solution: Let $u=3x+1$ then $du/dx=3$
$$u=3x+1 \implies 3x=u-1 \implies x=\frac{1}{3}(u-1) \implies x+1=\frac{1}{3}(u+2) $$
Now I get $$\frac{1}{3} \int (x+1)(3x+1)^9 (3 \,dx) = \frac{1}{3} \int \frac{1}{3}(u+2)u^9 du = \frac{1}{9} \int (u+2)u^9 du \\
= \frac{1}{9} \int (u^{10}+2u^9)\,du = \frac{1}{9}\left( \frac{u^{11}}{11}+\frac{2u^{10}}{10} \right) + C$$ 
But then I get to this one 
$$\int (x^2+2)(x-1)^7 dx$$ 
and the $x^2$ in brackets is throwing me off.
I put $u=x-1\implies x=u+1,$ hence $x^2+2 =(u+1)^2 +2 = u^2+3$. So
$$ \int(x^2+2)(x-1)^7\, dx = \int (u+1)u^7 du = \int (u+u^7) du = \frac{u^2}{2}+\frac{u^8}{8} + C $$
Is this correct or have I completely missed the point?

$\frac{19683 x^{11}}{11}+\frac{39366 x^{10}}{5}+15309 x^9+17496 x^8+13122 x^7+6804 x^6+\frac{12474 x^5}{5}+648 x^4+117 x^3+14 x^2+x$
