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I'm trying to find the integral of a function using the $\tan^{-1}$ trig sub method.

Function: $\displaystyle\frac{1}{z-Av^2}$ where $z$ and $A$ are constants.

To do the trig sub, I need to take the constants out to get the $\displaystyle\frac{1}{1+u^2}$ form (where $u$ would equal $v$ in this case). How do I do this?

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You let $u=\sqrt{\frac Az}v, v=\sqrt{\frac zA}u$ then $$\frac 1{z-Av^2}=\frac 1{z-zu^2}=\frac 1z\cdot \frac 1{1-u^2}$$ Note the sign on $u^2$ is different from what you asked, but a trig substitution still works.

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You can factorize the denominator and then do partial fractions assuming $ab>0$ $$I=\displaystyle\frac{1}{b-av^2}={\displaystyle\int}\dfrac{a}{\left(\sqrt{a}\sqrt{b}-ax\right)\left(ax+\sqrt{a}\sqrt{b}\right)}\,\mathrm{d}x$$ $$I=-{\displaystyle\int}\left(\dfrac{\sqrt{a}}{2\sqrt{b}\left(ax-\sqrt{a}\sqrt{b}\right)}-\dfrac{\sqrt{a}}{2\sqrt{b}\left(ax+\sqrt{a}\sqrt{b}\right)}\right)\mathrm{d}x$$

$$I=-\dfrac{\sqrt{a}}{2\sqrt{b}}{\displaystyle\int}\dfrac{1}{ax-\sqrt{a}\sqrt{b}}\,\mathrm{d}x+{\dfrac{\sqrt{a}}{2\sqrt{b}}}{\displaystyle\int}\dfrac{1}{ax+\sqrt{a}\sqrt{b}}\,\mathrm{d}x$$

$$I_1={\displaystyle\int}\dfrac{1}{ax-\sqrt{a}\sqrt{b}}\,\mathrm{d}x$$

Substitute $u=ax-\sqrt{a}\sqrt{b}$

$$I_1={\dfrac{1}{a}}{\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u=\dfrac{\ln\left(ax-\sqrt{a}\sqrt{b}\right)}{a}$$

$$I_2={\displaystyle\int}\dfrac{1}{ax+\sqrt{a}\sqrt{b}}\,\mathrm{d}x=\dfrac{\ln\left(ax+\sqrt{a}\sqrt{b}\right)}{a}$$

substitute the values of $I_1$,$I_2$ in $I$

$$I=-\dfrac{\ln\left(\frac{\left|ax-\sqrt{ab}\right|}{\left|ax+\sqrt{ab}\right|}\right)}{2\sqrt{ab}}+C$$

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