Question

Evaluate:

$\int \frac{x^3-4x^2+6x+5}{x^2-2x+3}dx$

Answer 1

$\int \frac{x^3-4x^2+6x+5}{x^2-2x+3}dx=\int (x-2-\frac{x-5}{x^2-2x+3})dx$

$=\int xdx-\int 2dx–\frac{1}{2}\int \frac{2x^{-}10}{x^2-2x+3}dx$

$=\frac{x^2}{2}-2x-\frac{1}{2}\int \frac{2x-2}{x^2-2x+3}+4\int dx{x}^2-2x+3$

$=\frac{x^2}{2}-2x-\frac{1}{2}\log (x^2-2x+3)+4\int dx{x}^2-2x+3$

$=\frac{x^2}{2}-2x-\frac{1}{2}\log (x^2-2x+3)+4\int dx{(x-1)}^2+\left({\sqrt{2}}^2\right)$

$=\frac{x^2}{2}-2x-\frac{1}{2}\log (x^2-2x+3)+4\int dx{(x-1)}^2+\left({\sqrt{2}}^2\right)$

$=\frac{x^2}{2}-2x-\frac{1}{2}\log (x^2-2x+3)+\frac{4}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1}{\sqrt{2}}\right)+C$.