# Integral question: $\int \frac{x^{5}+x+3}{x{^3}-5x^{2}}dx$

i have integral $\int \frac{x^{5}+x+3}{x{^3}-5x^{2}}dx$,

so first step is to divide polynoms, and i get:

$\frac{x^{5}+x+3}{x{^3}-5x^{2}}$ = $x^{2} + 5x + 25$ with remainder $125x^2 + x + 3$

is it corrent to divide this polynomial division into two integrals:

int 1: $\int x^{2}+5x+25 dx$

int 2: $\int \frac{125x^{2}+x+3}{x{^2}(x-5)}dx$

first integral solve directly from tables, and make partial fractions from second integral, then just merge solutions into one expression.

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Yes. It is correct. –  Beni Bogosel Sep 5 '11 at 19:21

## 3 Answers

Yes, it is all correct.

$$x^5 + x^2 + 3 = 125 x^2 + x + 3 + (x^3-5 x^2)(x^2 + 5 x + 25)$$ Hence $$\begin{eqnarray} \frac{x^5 + x + 3 }{x^3-5 x^2} &=& x^2 + 5 x + 25 + \frac{125 x^2 +x+ 3}{x^3-5 x} \\ &=& x^2 + 5 x + 25 -\frac{3}{5 \, x^2} - \frac{8}{25 x} + \frac{3133}{25} \frac{x - 5}{x^2-5} \end{eqnarray}$$

Therefore

$$\int \frac{x^5 + x + 3 }{x^3-5 x^2} \mathrm{d} x = \left(\frac{x^3}{3} + \frac{5}{2} x^2 + 25 x\right) + \left( \frac{3}{5} \frac{1}{x} - \frac{8}{25} \log x + \frac{3133}{25} \log(x-5) \right) + C$$

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@Thijs Yes, indeed, thank you. I will edit the post. –  Sasha Sep 5 '11 at 21:53

I liked the problem,so i decided to offer you a step by step solution.you may want to to go through this if you are after a step by step solution

As you can see the degree of the numerator is greater than the degree of the denominator you have divided using long division and have obtained an other expression which is

$$x^2 + 5x + 125 + \frac{125x^2 + x + 3}{x^2(x - 5)}$$

You must observe that the denominator is a product of a quadratic and a linear factor of the form

$$\frac{Ax + B}{ax^2 + bx +c} + \frac{C}{x + d}$$

Now use the Partial Fraction decomposition method to decompose the fraction into

$$\frac{-3}{5x^2} - \frac{8}{25x} + \frac{3133}{25(x-5)}$$

Now you can use the appropriate integral tricks to evaluate it.

Ok:I'am giving you the hint as per request.

$$\frac{Ax + B}{x^2} + \frac{C}{x-5}$$

Next step is to add the fraction after finding the common denominator.I think this should suffice!

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+1 But can you also give a hint about how you did this partial fraction decomposition? Thanks...(I noticed that your $\TeX$ing is better already, keep it up! :)) –  Srivatsan Sep 6 '11 at 3:37

Yes it is correct. Decompose integral 2 as partial fraction: $$\frac{125x^{2}+x+3}{x{^2}(x-5)} = -\frac{3}{5 x^2}-\frac{8}{25 x}+\frac{3133}{25 (x-5)}$$

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