x(x-y) Posted January 11, 2013 Share Posted January 11, 2013 I was just messing around with integrals to see if I could derive a general solution of this integral: [latex]I = \displaystyle \int \frac{dx}{\sqrt{c+bx+ax^2}}[/latex] and I was wondering if anybody could confirm my result. [latex]c+bx+ax^2 \Rightarrow c + a\left(x^2 + \frac{b}{a}x\right)[/latex] [latex]c+a\left[\left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right][/latex] [latex]\Rightarrow \left(c-\frac{b^2}{4a}\right)+a\left(x+\frac{b}{2a}\right)^2[/latex] [latex]\Rightarrow \displaystyle \int \frac{dx}{\sqrt{\left(c-\frac{b^2}{4a}\right)+a\left(x+\frac{b}{2a}\right)^2}}[/latex] [latex]\Rightarrow \frac{1}{\sqrt{c-\frac{b^2}{4a}}} \displaystyle \int \frac{dx}{\sqrt{1+\left(\sqrt{\frac{a}{c-\frac{b^2}{4a}}}\left(x+\frac{b}{2a}\right)\right)^2}}[/latex] then let [latex]\sinh t = \sqrt{\frac{a}{c-\frac{b^2}{4a}}}\left(x+\frac{b}{2a}\right)[/latex] [latex]\Rightarrow dx = \sqrt{\frac{c-\frac{b^2}{4a}}{a}} \hspace{2mm} d(\sinh t)[/latex] and we know that [latex]d(\sinh t) = \cosh t \; dt[/latex] So, substituting these terms into the integral gives [latex]\frac{1}{\sqrt{a}} \displaystyle \int \frac{\cosh t \; dt}{\sqrt{1 + \sinh^2 t}} \Rightarrow \frac{1}{\sqrt{a}} \displaystyle \int \frac{\cosh t \; dt}{\cosh t}[/latex] [latex]\Rightarrow \frac{1}{\sqrt{a}}t + C[/latex] [latex]\therefore I = \frac{1}{\sqrt{a}} \sinh^{-1} \left(\sqrt{\frac{a}{c-\frac{b^2}{4a}}}\left(x+\frac{b}{2a}\right)\right)+C[/latex] Link to comment Share on other sites More sharing options...
Bignose Posted January 11, 2013 Share Posted January 11, 2013 http://integral-table.com/integral-table.html#SECTION00004000000000000000 see eqn (39) also, you probably can check this yourself, right? By taking the derivative of that final expression and verifying that it results in the original integrand. Link to comment Share on other sites More sharing options...
x(x-y) Posted January 12, 2013 Author Share Posted January 12, 2013 http://integral-table.com/integral-table.html#SECTION00004000000000000000 see eqn (39) also, you probably can check this yourself, right? By taking the derivative of that final expression and verifying that it results in the original integrand. Thanks, and yeah I suppose I could've, oh well. Link to comment Share on other sites More sharing options...
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