Problem: integral of 1/(x*sqrt(x^(2)-4) dx
T tessec New member Joined Sep 21, 2006 Messages 4 Sep 24, 2006 #1 Problem: integral of 1/(x*sqrt(x^(2)-4) dx
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Sep 24, 2006 #2 hint ... \(\displaystyle \L \frac{d}{dx} [arcsec(x)] = \frac{1}{x\sqrt{x^2 - 1}}\)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Sep 24, 2006 #3 Hello, tessec! If you're familiar with Inverse Trig Functions, . . there is a formula for this one . . . \(\displaystyle \L\int\frac{dx}{x\sqrt{x^2\,-\,4}}\) Click to expand... Formula: \(\displaystyle \L\:\int\frac{du}{u\sqrt{u^2\,-\,a^2}} \;=\;\frac{1}{a}sec^{^{-1}}\left(\frac{u}{a}\right) \,+ \,C\)
Hello, tessec! If you're familiar with Inverse Trig Functions, . . there is a formula for this one . . . \(\displaystyle \L\int\frac{dx}{x\sqrt{x^2\,-\,4}}\) Click to expand... Formula: \(\displaystyle \L\:\int\frac{du}{u\sqrt{u^2\,-\,a^2}} \;=\;\frac{1}{a}sec^{^{-1}}\left(\frac{u}{a}\right) \,+ \,C\)
