can someone help me intergrate this.. i m most stuck on the long division part of it..

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

2. Hi

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

Hence $\frac{x^3-x^2+x-1}{x^2+9}=x-\frac{x^2+8x-1}{x^2+9}
=x-\frac{x^2+9+8x-10}{x^2+9}=x-1+\frac{8x}{x^2+9}-\frac{10}{x^2+9}$

3. Hello, Legendsn3verdie!

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

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

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

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