# Thread: One quick integral

1. ## One quick integral

I have seen this integral done in so many ways, and I am curious as to how each person (if anyone even does it) computes it on this site. I think it gives insight into that persons mathematical prefereneces. The integral is

$\int\sqrt{a+be^{cx}}dx$

Where $a,b,c$ are constants

2. I'd do $z^2=a+be^{cx}.$

3. Yeah me too, Here is my solution if anyone is interested

$\int\sqrt{a+be^{cx}}dx$

Let $\psi=\sqrt{a+be{cx}}\Rightarrow\frac{1}{c}\ln\left (\frac{\psi^2-a}{b}\right)=x$

So

$dx=\frac{2}{c}\frac{\psi}{\psi^2-a}d\psi$

So our integral turns into

$\frac{2}{c}\int\frac{\psi^2}{\psi^2-a}d\psi$

Now adding 0 we get

$\frac{2}{c}\bigg[\int\frac{\psi^2-a}{\psi^2-a}d\psi-a\int\frac{d\psi}{a-\psi^2}\bigg]$

Now for the second integral I rewrote this as

$a\frac{1}{a}\int\frac{d\psi}{1-\left(\frac{\psi}{\sqrt{a}}\right)^2}$

I then let $\varphi=\frac{\psi}{\sqrt{a}}\Rightarrow{d\psi=\sq rt{a}\varphi}$

So we have

$\sqrt{a}\int\frac{d\varphi}{1-\varphi^2}=\sqrt{a}{\rm{arctanh}}(\varphi)$

Back subbing we get

$\sqrt{a}{\rm{arctanh}}\left(\frac{\psi}{\sqrt{a}}\ right)$

So we have

$\frac{2}{c}\bigg[\psi-\sqrt{a}{\rm{arctanh}}\left(\frac{\psi}{\sqrt{a}}\ right)\bigg]$

Now for our final backsub we get

$\boxed{\int\sqrt{a+be^{cx}}dx=\frac{2}{c}\bigg[\sqrt{a+be^{cx}}-\sqrt{a}{\rm{arctanh}}\left(\frac{\sqrt{a+be^{cx}} }{\sqrt{a}}\right)\bigg]+C}$