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

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

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- Mar 11th 2010, 04:30 PMArchduke01Quick question involving long division
$\displaystyle \int \frac {x^3 + 13} {x^2 + 5x + 6} dx$

The first step is to do long division. Can anyone tell me if $\displaystyle (x+5) + \frac {-19x - 17} {x^2 + 5x + 6}$ is the right answer? - Mar 11th 2010, 04:32 PMMiss
- Mar 11th 2010, 04:43 PMArchduke01
- Mar 11th 2010, 05:09 PMArchie Meade
Hi Archduke,

you can check it quickly...

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

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

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

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

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

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

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