# Integrating a top heavy function

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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Instead of a substitution, try partial fractions. – mjh Apr 27 '14 at 17:08
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. – user124203 Apr 27 '14 at 17:08
when I did long devision, I got: \int x+10 + \frac{75x+250}{(x+5)^2}\ – user124203 Apr 27 '14 at 17:11
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 \$ ] . – RecklessReckoner Apr 27 '14 at 17:11

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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Much better than division followed by partial fractions, which I was going to post. – Mark Bennet Apr 27 '14 at 17:12
Right, :P ofcourse – user124203 Apr 27 '14 at 17:13
Thank you very much – user124203 Apr 27 '14 at 17:13
Why didn't the obvious elementary way occur to me before partial fractions? +1 – Sabyasachi Apr 27 '14 at 17:18
Because we see so many irreducible quadratics in the denominator? – RecklessReckoner Apr 27 '14 at 21:38

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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