Find the integral: $\int \frac{(x-6)^2}{x^4}\mathrm{d} x$

I have so far $\int (u)^2(u-6x)^{-4}\mathrm{d} x$

$u= x-6$ and $du=dx$ and $u-6=x$ Am I on the right track?


1 Answer 1


You don't need substitution. Simply expand $(x-6)^2$ and get $x^2 -12x +36$, and separate these term, i.e. $\frac{x^2 -12x +36}{x^4} = \frac{1}{x^2} - \frac{12}{x^3} + \frac{36}{x^4}$. This form will be familiar to you, integrate directly.

  • 2
    $\begingroup$ Makes Sense, however, I imagine it's not reasonable strategy if the power is greater than 2? $\endgroup$
    – Cetshwayo
    Apr 26, 2015 at 20:49
  • $\begingroup$ It's still a reasonable strategy as there is no other strategy $\endgroup$ Apr 26, 2015 at 20:55
  • $\begingroup$ Once the denominator is $x^n$, this strategy provides solution. $\endgroup$
    – MonkeyKing
    Apr 26, 2015 at 20:56
  • $\begingroup$ @MonkeyKing apologies for downvoting you, but my cat accidentally stepped on the downvote button (on my iPad). So to make amends I better upvote you. $\endgroup$ Apr 26, 2015 at 20:58

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