How do you integrate $int (2x-3)/(x^2-3x+5)$ using substitution?

Question

10268 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-17T14:14:02

$int (2x-3)/(x^2-3x+5)dx = ln (x^2-3x+5) + C$

Note that:

$d/dx (x^2-3x+5) = 2x-3$

So if we substitute:

$t = x^2-3x+5$

$dt = (2x-3)dx$

we have:

$int (2x-3)/(x^2-3x+5)dx = int (dt)/t = ln abs(t) +C = ln (x^2-3x+5) + C$

How do you integrate int (2x-3)/(x^2-3x+5)int (2x-3)/(x^2-3x+5) using substitution?