$$ \int \frac { 3 ^ { x } } { 27 ^ { x } - 1 } d x $$

Question

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To evaluate the integral:

$$
\int \frac{3^x}{27^x - 1} \, dx
$$

we can follow these steps:

### Step 1: Simplify the Integrand
First, recognize that $$27^x = (3^3)^x = 3^{3x}$$. Therefore, the integral becomes:

$$
\int \frac{3^x}{3^{3x} - 1} \, dx = \int \frac{3^x}{3^{3x} - 1} \, dx
$$

### Step 2: Substitution
Let $$u = 3^x$$. Then, the derivative of $$u$$ with respect to $$x$$ is:

$$
\frac{du}{dx} = \ln(3) \cdot 3^x \implies dx = \frac{du}{\ln(3) \cdot u}
$$

Substituting these into the integral:

$$
\int \frac{u}{u^3 - 1} \cdot \frac{1}{\ln(3) \cdot u} \, du = \frac{1}{\ln(3)} \int \frac{1}{u^3 - 1} \, du
$$

### Step 3: Partial Fraction Decomposition
Factor the denominator:

$$
u^3 - 1 = (u - 1)(u^2 + u + 1)
$$

Express the integrand as partial fractions:

$$
\frac{1}{u^3 - 1} = \frac{A}{u - 1} + \frac{Bu + C}{u^2 + u + 1}
$$

Solving for $$A$$, $$B$$, and $$C$$, we find:

$$
A = \frac{1}{3}, \quad B = -\frac{1}{3}, \quad C = -\frac{2}{3}
$$

Thus,

$$
\frac{1}{u^3 - 1} = \frac{1/3}{u - 1} + \frac{-\frac{1}{3}u - \frac{2}{3}}{u^2 + u + 1}
$$

### Step 4: Integrate Each Term
Now, integrate term by term:

$$
\frac{1}{\ln(3)} \int \left( \frac{1/3}{u - 1} + \frac{-\frac{1}{3}u - \frac{2}{3}}{u^2 + u + 1} \right) \, du
$$

This splits into:

$$
\frac{1}{3 \ln(3)} \ln|u - 1| - \frac{1}{6 \ln(3)} \ln(u^2 + u + 1) - \frac{\sqrt{3}}{3 \ln(3)} \arctan\left(\frac{2u + 1}{\sqrt{3}}\right) + C
$$

### Step 5: Substitute Back $$u = 3^x$$
Finally, substitute $$u = 3^x$$ back into the expression:

$$
\begin{aligned}
\int \frac{3^x}{27^x - 1} \, dx &= \frac{1}{3 \ln(3)} \ln|3^x - 1| - \frac{1}{6 \ln(3)} \ln(3^{2x} + 3^x + 1) \
&\quad - \frac{\sqrt{3}}{3 \ln(3)} \arctan\left(\frac{2 \cdot 3^x + 1}{\sqrt{3}}\right) + C
\end{aligned}
$$

where $$C$$ is the constant of integration.

### Final Answer

After expanding, an antiderivative is (1⁄3 ln3) ln (3ˣ – 1) minus one sixth ln(3²ˣ + 3ˣ + 1) minus √3⁄3 arctan[(2·3ˣ + 1)/√3] plus C. Thus,

∫3^x/(27^x−1)dx = (1/(3 ln3)) ln(3^x−1) − (1/(6 ln3)) ln(3^{2x}+3^x+1) − (√3/3 ln3) arctan((2·3^x+1)/√3) + C
The integral of $$ \frac{3^x}{27^x - 1} $$ with respect to $$ x $$ is: $$ \frac{1}{3 \ln 3} \ln(3^x - 1) - \frac{1}{6 \ln 3} \ln(3^{2x} + 3^x + 1) - \frac{\sqrt{3}}{3 \ln 3} \arctan\left(\frac{2 \cdot 3^x + 1}{\sqrt{3}}\right) + C $$ where $$ C $$ is the constant of integration.

\( \int \frac { 3 ^ { x } } { 27 ^ { x } - 1 } d x \)

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