Evaluating Integral: \int\frac{4}{x^2-1}

[danielatha4]

Homework Statement

$$\int$$$$\frac{4}{x^2-1}$$

Homework Equations

The Attempt at a Solution

I thought I had this right...

$$\frac{4}{x^2-1}$$ = 2 ($$\frac{1}{x-1}$$-$$\frac{1}{x+1}$$)

therefore,

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

I then have to evaluate from 2 to 3 and I get .814, but it isn't right.

 

[danielatha4]

Nevermind, I believe I figured it out.

Edit: Okay, I don't understand where I went wrong. Any help please?

 

[Bohrok]

I get log(9/4) ≈ 0.81093
Perhaps you entered it wrong in a calculator?

 

[danielatha4]

Oh gees... rounding error... thanks