Different results from the same integral with two different methods

Question

$$\int x^2(x-3)^{11}\,dx$$

By substituting (let $t=(x-3)$), it results in one answer and integration by parts (let $u=x^2$ and $v=(x-3)^{11}$), results in something that is totally different from the first.

by substitution method

let $t=(x-3)$ then $x^2=(t^2+6t+9)$

$\int x^2(x-3)^{11}\,dx=\int (t^2+6t+9)t^{11}\,dt$

$=\frac{t^{14}}{14}+\frac{6t^{13}}{13}+\frac{3t^{12}}4 +C$

$=\frac{(x-3)^{14}}{14}+\frac{6(x-3)^{13}}{13}+\frac{3(x-3)^{12}}4 +C$

and by Integration by parts

$\int x^2(x-3)^{11}\,dx$

$=\frac{x^2(x-3)^{12}}{12}-\int\frac{(x-3)^{12}}{12}(2x)dx$

$=\frac{x^2(x-3)^{12}}{12}-\frac{x(x-3)^{13}}{78}+\frac{(x-3)^{14}}{1092} +C$

these two solutions are different from each other.

Answer 1

Those two results are actually the same, as you can see here:

http://www.wolframalpha.com/input/?i=%28x-3%29%5E14%2F14%2B6%28x-3%29%5E13%2F13%2B3%28x-3%29%5E12%2F4

http://www.wolframalpha.com/input/?i=x%5E2%28x-3%29%5E12%2F12-x%28x-3%29%5E13%2F78%2B%28x-3%29%5E14%2F1092

in the "Expanded form" entry.

Answer 2

The first:

$\dfrac{(x-3)^{14}}{14}+\dfrac{6(x-3)^{13}}{13}+\dfrac{3(x-3)^{12}}{4}+C=(x-3)^{12}\left[\dfrac{(x-3)^{2}}{14}+\dfrac{6(x-3)}{13}+\dfrac{3}{4}\right]+C=(x-3)^{12}\left[\dfrac{x^2}{14}-\dfrac{6x}{14}+\dfrac{9}{14}+\dfrac{6x}{13}-\dfrac{18}{13}+\dfrac{3}{4}\right]+C=(x-3)^{12}\left[\dfrac{x^2}{14}+\dfrac{3x}{91}+\dfrac{3}{364}\right]+C$

The second:

$\dfrac{x^2(x-3)^{12}}{12}-\dfrac{x(x-3)^{13}}{78}+\dfrac{(x-3)^{14}}{1092} +C=(x-3)^{12}\left[\dfrac{x^2}{12}-\dfrac{x^2}{78}+\dfrac{3x}{78}+\dfrac{x^2}{1092}-\dfrac{6x}{1092}+\dfrac{9}{1092}\right]+C=(x-3)^{12}\left[\dfrac{x^2}{14}+\dfrac{3x}{91}+\dfrac{3}{364}\right]+C$