Integral of [ 1 / ( 1 - x^2) ] Is the only way to do this problem by using integration by parts?
G grapz Junior Member Joined Jan 13, 2007 Messages 80 Mar 3, 2008 #1 Integral of [ 1 / ( 1 - x^2) ] Is the only way to do this problem by using integration by parts?
D Dr. Flim-Flam Junior Member Joined Oct 10, 2007 Messages 108 Mar 3, 2008 #2 Re: Integral of 1 / 1 - x ^2 Let x = sin[theta], then dx = cos[theta]d[theta]. Hence integral[1/(1-x^{2})] = ln|sec[theta]+tan[theta] = (1/2) ln|(x+1)/(x-1)|. Note: I omitted a lot of steps for brevity,but can you take it from here?
Re: Integral of 1 / 1 - x ^2 Let x = sin[theta], then dx = cos[theta]d[theta]. Hence integral[1/(1-x^{2})] = ln|sec[theta]+tan[theta] = (1/2) ln|(x+1)/(x-1)|. Note: I omitted a lot of steps for brevity,but can you take it from here?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 3, 2008 #3 Hello, grapz! \(\displaystyle \int \frac{dx}{1 - x^2}\) Is the only way to do this problem by using integration by parts? Click to expand... By parts? . .. I think you mean Partial Fractions. The Good Doctor is right . . . Trig subtitution is a good method. When I took Calculus (during the Jurassic Period), . . I memorized two more formulas: . . . . \(\displaystyle \int\frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+ C\) . . . .\(\displaystyle \int\frac{du}{a^2-u^2} \;=\;\frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right| + c\) so this problem was a piece of cake.
Hello, grapz! \(\displaystyle \int \frac{dx}{1 - x^2}\) Is the only way to do this problem by using integration by parts? Click to expand... By parts? . .. I think you mean Partial Fractions. The Good Doctor is right . . . Trig subtitution is a good method. When I took Calculus (during the Jurassic Period), . . I memorized two more formulas: . . . . \(\displaystyle \int\frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+ C\) . . . .\(\displaystyle \int\frac{du}{a^2-u^2} \;=\;\frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right| + c\) so this problem was a piece of cake.
