Stuck on an indefinite integral probably using hyperbolic substitution.

Question

First off, please don't give the answer. I'm really after a starting point.

I'm trying to solve the integral

$$\int \frac{1}{25e^x+9}~dx$$

I have done a few others where I have an $x$ instead of an $e^x$ and it's under a root. To solve those I've been using $sinh(x)$ or $cosh(x)$ substitutions so in this case I imagine I should be trying $tanh(x)$. But not sure, the $e^x$ is throwing me off. It's not squared, and it is $e^x$ not $x$.

Any guidance on how to tackle this would be much appreciated.

Thanks

Answer 1

Hint: write $$\frac{1}{25 e^x + 9} = - \frac 1 9 \frac{-9e^{-x}}{25 + 9 e^{-x}}$$

Answer 2

$$\frac1{25e^x+9}=\frac{e^{-x}}{25+9e^{-x}}$$

Set $25+9e^{-x}=u$

---

Alternatively, setting $25e^x+9=v,25e^x\ dx=dv\iff dx=\dfrac{dv}{v-9}$

$$\implies I=\int\frac{dx}{25e^x+9}=\int\frac{dv}{v(v-9)}$$

$$\implies 9I=\int\frac{v-(v-9)}{v(v-9)}dv=\cdots$$