Solving $\int \frac{1}{\sqrt{1-a+ax^2}}dx $ This is part of a large engineering report and through plugging this into wolfram alpha I have an answer that, once computed, works perfectly for what I need. However, I am struggling to see how to go from the input to the output, if anyone could point me in the right direction I would be grateful.
$ \int \frac{1}{\sqrt{1-a+ax^2}}dx = \frac{1}{\sqrt{a}} ln(\sqrt{a}\sqrt{a(x^2 - 1) +1} +ax)
 $
 A: By a suitable scaling, we can reduce this to "well-known" integrals
$$ \int \dfrac{dt}{\sqrt{1-t^2}} = \arcsin(t) + C$$ 
or 
$$ \int \dfrac{dt}{\sqrt{1+t^2}} = \text{arcsinh}(t) + C = \ln\left(t + \sqrt{1+t^2}\right)+C $$
(depending on the sign of $a/(1-a)$.
If they are not well-known enough for you, try the substitution
$t = \sin(\theta)$ or $t = \sinh(\theta)$ respectively.
A: We have the more general problem:
$$\int\frac1{\sqrt{a+bx^2}}\ dx$$
The general trick would then to factor out $\sqrt a$, and let $c^2=b/a$, assuming $a,b$ to be positive.
$$\frac1{\sqrt a}\int\frac1{\sqrt{1+(cx)^2}}\ dx$$
And now let $\tan u=cx$,
$$\frac1{c\sqrt a}\int\frac{\sec^2u}{\sqrt{1+\tan^2u}}\ du=\frac1{c\sqrt a}\int\frac{\sec^2u}{\sec u}\ du=\frac1{c\sqrt a}\int\sec u\ du$$
And that last bit may be handled with integration by parts, Google, or other nefariously popular method.
In the event that $a$ or $b$ are negative, then we simply use $\arcsin$.
$$\arcsin(x)=\int\frac1{\sqrt{1-x^2}}\ dx$$
or one may recall $\operatorname{arcsinh}$ instead of all the above:
$$\operatorname{arcsinh}(x)=\int\frac1{\sqrt{1+x^2}}\ dx$$
and then use identites to reach your form.
