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How do you integrate this function?

$$\int\frac{x^3}{(x+5)^2}dx$$ I have tried it myself by substitution but I can't seem to get rid of the $x$s.

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    $\begingroup$ Instead of a substitution, try partial fractions. $\endgroup$ – mjh Apr 27 '14 at 17:08
  • $\begingroup$ tried as well but since the bottom has (x+5) twice, A and B will cancel meaning I can't find a value for them. $\endgroup$ – user124203 Apr 27 '14 at 17:08
  • $\begingroup$ when I did long devision, I got: \int x+10 + \frac{75x+250}{(x+5)^2}\ $\endgroup$ – user124203 Apr 27 '14 at 17:11
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    $\begingroup$ You need to make a polynomial division first, since the partial fractions method requires the degree of the polynomial in the numerator to be lower that that in the denominator. [OK, in your division, the quotient should be $ \ x - 10 \ $ ] . $\endgroup$ – colormegone Apr 27 '14 at 17:11
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Hint: Rewrite it as $$\int \frac{(u-5)^3}{u^2}\text{d}u$$ by making the substitution $u = x+5$ and rearranging that until you can get $x$ (i.e., if $u = x+5$, then $x = ?$)

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  • $\begingroup$ Much better than division followed by partial fractions, which I was going to post. $\endgroup$ – Mark Bennet Apr 27 '14 at 17:12
  • $\begingroup$ Right, :P ofcourse $\endgroup$ – user124203 Apr 27 '14 at 17:13
  • $\begingroup$ Why didn't the obvious elementary way occur to me before partial fractions? +1 $\endgroup$ – Guy Apr 27 '14 at 17:18
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    $\begingroup$ Because we see so many irreducible quadratics in the denominator? $\endgroup$ – colormegone Apr 27 '14 at 21:38
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Long division: $$ \begin{array}{cccccc} & & x & - & 10 \\ \\ x^2+10x+25 & ) & x^3 \\ & & x^3 & + & 10x^2 & + & 25x \\ \\ & & & & -10x^2 & - & 25x \\ & & & & -10x^2 & - & 100x \\ \\ & & & & & & 75 x \end{array} $$

So we have $$ \frac{x^3}{x^2+10x+25} = x - 10 + \frac{75x}{(x+5)^2} = x - 10 + \frac{A}{x-5} + \frac{B}{(x+5)^2}. $$

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