`int(1)/(3+(2-3x)^(2)) dx`

Answer

To solve the integral \( \int \frac{1}{3 + (2 - 3x)^2} \, dx \), we can follow these steps:

### Step 1: Identify the form of the integral
We recognize that the integral resembles the form \( \int \frac{1}{a^2 + u^2} \, du \), where \( a = \sqrt{3} \) and \( u = 2 - 3x \).

### Step 2: Rewrite the integral
We can rewrite the integral as:
\[ \int \frac{1}{3 + (2 - 3x)^2} \, dx = \int \frac{1}{\sqrt{3}^2 + (2 - 3x)^2} \, dx \]

### Step 3: Substitute for \( u \)
Let \( u = 2 - 3x \). Then, differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = -3 \quad \Rightarrow \quad dx = \frac{du}{-3} \]

### Step 4: Substitute \( dx \) in the integral
Substituting \( u \) and \( dx \) into the integral gives:
\[ \int \frac{1}{\sqrt{3}^2 + u^2} \cdot \frac{du}{-3} = -\frac{1}{3} \int \frac{1}{\sqrt{3}^2 + u^2} \, du \]

### Step 5: Apply the formula for integration
Using the formula \( \int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \), we find:
\[ -\frac{1}{3} \cdot \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{u}{\sqrt{3}} \right) + C = -\frac{1}{3\sqrt{3}} \tan^{-1} \left( \frac{u}{\sqrt{3}} \right) + C \]

### Step 6: Substitute back for \( u \)
Substituting back \( u = 2 - 3x \):
\[ -\frac{1}{3\sqrt{3}} \tan^{-1} \left( \frac{2 - 3x}{\sqrt{3}} \right) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \frac{1}{3 + (2 - 3x)^2} \, dx = -\frac{1}{3\sqrt{3}} \tan^{-1} \left( \frac{2 - 3x}{\sqrt{3}} \right) + C \]