How do you find the antiderivative of dx/(e^(2x)-9)^(1/2)?

1 Answer
Feb 19, 2015

The answer is: $\frac{1}{3} \arctan \left(\frac{\sqrt{{e}^{2 x} - 9}}{3}\right) + c$.

We have to make this substitution:

$\sqrt{{e}^{2 x} - 9} = t \Rightarrow {e}^{2 x} - 9 = {t}^{2} \Rightarrow {e}^{2 x} = {t}^{2} + 9 \Rightarrow$

$2 x = \ln \left({t}^{2} + 9\right) \Rightarrow x = \frac{1}{2} \ln \left({t}^{2} + 9\right) \Rightarrow \mathrm{dx} = \frac{1}{2} \cdot \frac{2 t}{{t}^{2} + 9} \mathrm{dt} \Rightarrow$

$\mathrm{dx} = \frac{t}{{t}^{2} + 9} \mathrm{dt}$,

so:

$\int \frac{\mathrm{dx}}{\sqrt{{e}^{2 x} - 9}} = \int \frac{1}{t} \cdot \frac{t}{{t}^{2} + 9} \mathrm{dt} = \int \frac{\mathrm{dt}}{{t}^{2} + 9} =$

$\int \frac{\mathrm{dt}}{9 \left({t}^{2} / 9 + 1\right)} = \frac{1}{9} \int \frac{\mathrm{dt}}{{\left(\frac{t}{3}\right)}^{2} + 1} = \frac{1}{9} \cdot 3 \int \frac{\frac{1}{3}}{{\left(\frac{t}{3}\right)}^{2} + 1} \mathrm{dt} =$

$= \frac{1}{3} \arctan \left(\frac{t}{3}\right) + c = \frac{1}{3} \arctan \left(\frac{\sqrt{{e}^{2 x} - 9}}{3}\right) + c$.