Indefinite Integral (Inverse trigonometric functions)

Question

How to evaluate this given expression? $$\int\frac{du}{\sqrt{9e^{-2u}-1}}$$ I got so many tries but I'm not sure of my answer because somebody said that it was wrong, they told me that I used a wrong formula applied! That's why I ask a support here I want correct explanation and answer of this given!

Thanks!

Answer 1

try to substitute $\frac{1}{3}e^{u} = cost$ then $du = -tg(t)dt$. And the integral will be $\int \frac{-tgtdt}{\sqrt{\frac{1}{cos(t)^2} - 1}}$ then use that $(cos(t))^2 + (sin(t))^2 = 1$

Answer 2

Considering $$I=\int\frac{du}{\sqrt{9e^{-2u}-1}}$$ use $$\sqrt{9e^{-2u}-1}=t\implies u=-\frac{1}{2} \log \left(\frac{1}{9} \left(t^2+1\right)\right)\implies du=-\frac{t}{t^2+1}\,dt$$ This makes $$I=-\int\frac{dt}{t^2+1}$$