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

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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.
 
Last edited:
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Nevermind, I believe I figured it out.

Edit: Okay, I don't understand where I went wrong. Any help please?
 
Last edited:
I get log(9/4) ≈ 0.81093
Perhaps you entered it wrong in a calculator?
 
Oh gees... rounding error... thanks
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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