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]

 

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.

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.