Question

Solve:

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

Answer 1

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

This can aslo be written as

$=\int x-2+\frac{11-x}{x^2-2x+3}dx$

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

$=\frac{x^2}{2}-2x+\frac{10}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1}{\sqrt{2}}\right)-\frac{1}{2}\log |x^2-2x+3|+c$