Evaluate the integral using any appropriate algebraic method or trigonometric identity.

Ask by Carter Norris. in the United States

Jan 21,2025

Question

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

UpStudy AI · Accepted Answer

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.

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