How to solve $\int\frac{2+3x}{3x^2+4x-6}\,dx$?

Question

I am trying to solve $$\int\frac{2+3x}{3x^2+4x-6}\,dx$$ using partial fraction decomposition but it fails, how to solve it?

Answer 1

Substitute $$u= 3x^2+4x-6$$

Then $$\frac {du}{dx}= 6x+4$$

$$I=\int\frac{2+3x}{3x^2+4x-6}\,dx=\frac 12\int \frac { du}u=\frac {\ln|u |}2+K$$ $$I=\frac {\ln|3x^2+4x-6|}2+K$$

Answer 2

$\int\frac{2+3x}{3x^2+4x-6}=\frac{1}{2}\int\frac{6x+4}{3x^2+4x-6}=\frac{\ln(|3x^2+4x-6|)}{2}+K$

Since $6x+4$ is the derivative of $3x^2+4x-6$