Quick question involving long division

**Archduke01** · March 11th 2010, 04:30 PM

$\int \frac {x^3 + 13} {x^2 + 5x + 6} dx$

The first step is to do long division. Can anyone tell me if $(x+5) + \frac {-19x - 17} {x^2 + 5x + 6}$ is the right answer?

**Miss** · March 11th 2010, 04:32 PM

Originally Posted by Archduke01

$\int \frac {x^3 + 13} {x^2 + 5x + 6} dx$

The first step is to do long division. Can anyone tell me if $(x+5) + \frac {-19x - 17} {x^2 + 5x + 6}$ is the right answer?

If you make a common denominator, do you will back to the original fraction?

**Archduke01** · March 11th 2010, 04:43 PM

Originally Posted by Miss

If you make a common denominator, do you will back to the original fraction?

Sorry? I don't understand your question.

**Archie Meade** · March 11th 2010, 05:09 PM

Originally Posted by Archduke01

$\int \frac {x^3 + 13} {x^2 + 5x + 6} dx$

The first step is to do long division. Can anyone tell me if $(x+5) + \frac {-19x - 17} {x^2 + 5x + 6}$ is the right answer?

Hi Archduke,

you can check it quickly...

$x^3+13=\left(x^2+5x+6\right)(x+a)+Rem$

$=x^3+5x^2+6x+ax^2+5ax+6a+Rem$

$=x^3+(5+a)x^2+(6+5a)x+6a+Rem$

Therefore $a=-5$ as $x^3+13$ has no $x^2$ term.

It also has no x term, so we cancel it in the remainder

$x^3+13=x^3-19x-30+19x+43$

$\frac{x^3+13}{x^2+5x+6}=(x-5)+\frac{19x+43}{x^2+5x+6}$

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