Evaluate the integral using any appropriate algebraic method or trigonometric identity. $$ \int \frac{d \mathrm{x}}{25 e^{-9 \mathrm{x}}+e^{9 x}} $$

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To evaluate the integral

$$
\int \frac{dx}{25 e^{-9x} + e^{9x}},
$$

let's proceed with a substitution to simplify the expression.

**Step 1: Substitution**

Let $$ t = e^{9x} $$. Then, the differential $$ dx $$ can be expressed in terms of $$ dt $$:

$$
dt = 9 e^{9x} dx \quad \Rightarrow \quad dx = \frac{dt}{9t}.
$$

**Step 2: Rewrite the Integral**

Substituting $$ t $$ and $$ dx $$ into the integral:

$$
\int \frac{dx}{25 e^{-9x} + e^{9x}} = \int \frac{1}{25 \frac{1}{t} + t} \cdot \frac{dt}{9t}.
$$

Simplifying the denominator:

$$
25 \frac{1}{t} + t = \frac{25 + t^2}{t}.
$$

Thus, the integral becomes:

$$
\int \frac{t}{25 + t^2} \cdot \frac{dt}{9t} = \frac{1}{9} \int \frac{dt}{25 + t^2}.
$$

**Step 3: Integration**

The integral $$ \int \frac{dt}{25 + t^2} $$ is a standard arctangent form:

$$
\int \frac{dt}{a^2 + t^2} = \frac{1}{a} \arctan\left(\frac{t}{a}\right) + C,
$$

where $$ a = 5 $$ in this case.

Therefore,

$$
\frac{1}{9} \int \frac{dt}{25 + t^2} = \frac{1}{9} \cdot \frac{1}{5} \arctan\left(\frac{t}{5}\right) + C = \frac{1}{45} \arctan\left(\frac{e^{9x}}{5}\right) + C.
$$

**Final Answer:**

$$
\boxed{\frac{1}{45}\, \tan^{-1}\left(\frac{\,e^{9x}\,}{\,5\,}\right) + C}
$$
The integral evaluates to $$ \frac{1}{45} \arctan\left(\frac{e^{9x}}{5}\right) + C $$.

Evaluate the integral using any appropriate algebraic method or trigonometric identity. \( \int \frac{d \mathrm{x}}{25 e^{-9 \mathrm{x}}+e^{9 x}} \)

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