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

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

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


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

these two solutions are different from each other.

  • 2
    $\begingroup$ Could you show what you did and what you obtained ? $\endgroup$ Dec 27, 2014 at 16:27
  • $\begingroup$ Show a bit of work so we can help find your mistake. Those two methods, performed correctly, should yield the same answer. $\endgroup$
    – 123
    Dec 27, 2014 at 16:37
  • $\begingroup$ Conveniently, this is a polynomial, and therefor extremely straightforward to check. You could multiply it all out. Or you could check your two answers by differentiating them, and seeing if they are the same polynomial as the integrand. Or you could ask W|A if you just want a correct answer to check against. $\endgroup$
    – davidlowryduda
    Dec 27, 2014 at 16:37
  • $\begingroup$ Did you substitute $x=t+3$ back afterwards? $\endgroup$
    – copper.hat
    Dec 27, 2014 at 16:37
  • $\begingroup$ Sir the question is edited. $\endgroup$
    – Ahmad
    Dec 27, 2014 at 17:05

2 Answers 2


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



in the "Expanded form" entry.


The first:


The second:

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


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