Integrating $\int \ln(x^2-2x-3)\,dx$

Question

I need to integrate this:

$$\int \ln\left(x^2-2x-3\right)\,dx$$

I'm thinking of going this direction:

$$\int \ln\left(x^2-2x-3\right)\,dx=\int \ln((x-3)(x+1))\,dx$$

Which would then equal $$\int \ln(x-3)\,dx + \int \ln(x+1)\,dx$$

What do you think? I guess I could also go this way: $$u=\ln\left(x^2-2x-3\right) \; \; \; dv=1\,dx$$ meaning I could integrate by parts. Which way is the best way? Can I only choose one?

Answer 1

For $x>3$ we obtain: $$\int\ln(x^2-2x-3)dx=\int\ln(x+1)dx+\int\ln(x-3)dx=$$ $$=(x+1)\ln(x+1)-(x+1)+(x-3)\ln(x-3)-(x-3)+C=$$ $$=(x+1)\ln(x+1)+(x-3)\ln(x-3)-2x+C.$$

Answer 2

Beside to Michael answer, you may use integration by parts, i.e., by taking

$$\ln(x^2-2x-3)=u, dx=dv$$ over a valid interval and so it leads you to do a partial fraction way for the rest.

Answer 3

Integrating by parts: $$F(x)=x\ln{[(x+1)(x-3)]}-\int x \cdot \frac{2x-2}{(x+1)(x-3)}dx = x\ln{[(x+1)(x-3)]} - \int \left(2-\frac{1}{x+1}+\frac{3}{x-3}\right)dx$$ Can you finish?