$\begin{array}{l}\text{The integral}\int \frac{e^{-x}}{9+4e^{-2x}}dx\text{ is equal to}\end{array}$

The correct answer is $\frac{-1}{6}{tan}^{-1}\left(\frac{{\mathrm{2e}}^{-x}}{3}\right)+C$

We need to evaluate the integral:

$I=\int \frac{e^{-x}}{9+4e^{-2x}}dx$

Step 1: Choose a suitable substitution.

Notice that the numerator contains $e^{-x}$ and the denominator contains $e^{-2x}={\left(}^{e^{-x}}$. This strongly suggests the substitution:

$t=e^{-x}$

Differentiating both sides with respect to x:

$\frac{\mathrm{dt}}{\mathrm{dx}}=-e^{-x}=-t$

Therefore:

$\mathrm{dt}=-e^{-x}\mathrm{dx}⟹e^{-x}\mathrm{dx}=-\mathrm{dt}$

Step 2: Substitute into the integral.

Replacing $e^{-x}\mathrm{dx}$ with $-\mathrm{dt}$ and $e^{-2x}$ with $t^2$:

$I=\int \frac{-\mathrm{dt}}{9+4t^2}$

Step 3: Factor the denominator to match the standard arctangent form.

Recall the standard integration formula:

$\int \frac{\mathrm{dt}}{a^2+t^2}=\frac{1}{a}{tan}^{-1}\left(\frac{t}{a}\right)+C$

Factor out 4 from the denominator so it looks like $a^2+t^2$:

$9+4t^2=4\left(\frac{9}{4}+t^2\right)=4\left({\left(\frac{3}{2}\right)}^2+t^2\right)$

So the integral becomes:

$I=-\int \frac{\mathrm{dt}}{4\left({\left(\frac{3}{2}\right)}^2+t^2\right)}=-\frac{1}{4}\int \frac{\mathrm{dt}}{{\left(\frac{3}{2}\right)}^2+t^2}$

Step 4: Apply the standard arctangent formula.

Here $a=\frac{3}{2}$. Applying the formula:

$I=-\frac{1}{4}·\frac{1}{\frac{3}{2}}{tan}^{-1}\left(\frac{t}{\frac{3}{2}}\right)+C$

Simplify $\frac{1}{\frac{3}{2}}=\frac{2}{3}$, and $\frac{t}{\frac{3}{2}}=\frac{2}{3}t$:

$I=-\frac{1}{4}·\frac{2}{3}{tan}^{-1}\left(\frac{2}{3},t\right)+C$

$I=-\frac{2}{12}{tan}^{-1}\left(\frac{2}{3},t\right)+C=-\frac{1}{6}{tan}^{-1}\left(\frac{2}{3},t\right)+C$

Step 5: Back-substitute $t=e^{-x}$.

Replace $t$ with $e^{-x}$:

$I=-\frac{1}{6}{tan}^{-1}\left(\frac{2e^{-x}}{3}\right)+C$

Why the answer is negative: The key reason the coefficient is $\frac{-1}{6}$ and not $\frac{1}{6}$ is because when we differentiated the substitution $t=e^{-x}$, we obtained $\mathrm{dt}=-e^{-x}\mathrm{dx}$, introducing a minus sign that propagates through the entire calculation.

Therefore, the final answer is:

$I=\frac{-1}{6}{tan}^{-1}\left(\frac{2e^{-x}}{3}\right)+C$