• Apr 27th 2008, 01:34 PM
Legendsn3verdie
can someone help me intergrate this.. i m most stuck on the long division part of it..

$\displaystyle \int \frac{x^3 -x^2 +x-1}{x^2 +9} dx$
• Apr 27th 2008, 01:46 PM
flyingsquirrel
Hi

$\displaystyle x^3-x^2+x-1=x(x^2+9)-9x-x^2+x-1=x(x^2+9)-x^2-8x-1$ (the only way you can build $\displaystyle x^3$ using $\displaystyle x^2+9$ is multiplying $\displaystyle x$ by $\displaystyle x^2+9$)

Hence $\displaystyle \frac{x^3-x^2+x-1}{x^2+9}=x-\frac{x^2+8x-1}{x^2+9} =x-\frac{x^2+9+8x-10}{x^2+9}=x-1+\frac{8x}{x^2+9}-\frac{10}{x^2+9}$
• Apr 27th 2008, 02:31 PM
Soroban
Hello, Legendsn3verdie!

Quote:

$\displaystyle \int \frac{x^3 -x^2 +x-1}{x^2 +9}\,dx$

$\displaystyle \begin{array}{cccccccccc}& & & & & &x & \text{-} & 1 \\& & - & -& - & - & - & - & - \\x^2+9 & | & x^3 & \text{-} &x^2 & + & x & \text{-} & 1 \\& & x^3 & & & + & 9x \\& & -& -& -& -& - \\& & & \text{-} & x^2 & \text{-} & 8x &\text{-}& 1 \\ & & & \text{-} & x^2 & & & \text{-} & 9 \\ & & &- & -& -& - &-& - \\ & & & & & \text{-} & 8x & + & 8\end{array}$

Hence: . $\displaystyle \frac{x^3-x^2+x-1}{x^2+9} \;\;=\;\; x - 1 + \frac{\text{-}8x+8}{x^2+9} \;\;=\;\;x - 1 - \frac{8x}{x^2+9} + \frac{8}{x^2+9}$

Therefore: .$\displaystyle \int\frac{x^3-x^2+x-1}{x^2+9}\,dx \;\;=\;\;\int(x-1)\,dx \:-\: 8\!\int\!\frac{x}{x^2+9}\,dx \:+ \:8\!\int\!\frac{dx}{x^2+9}$