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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) $

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    How much integral calculus do you know? Have you worked with integrals before? – 2017-01-24
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    I am a third year engineering student, I've done plenty of integral work before so I should be able to follow whatever you suggest. – 2017-01-24
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    Hint: substitute $x=\sqrt{1-a}y / \sqrt a$ – 2017-01-24
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    OK, cool. If no one has tackled this by tomorrow morning I'll get it for you. For the moment, at least check that your answer's derivative is your integrand. Doing this check should give you some sense of how to go in the other direction. – 2017-01-24
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    Futhermore, do you know the derivative of $\arcsin$? – 2017-01-24

2 Answers 2

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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.

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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.