Find integrals of given functions
`int_(0)^(1) (dx)/(3x + 2)`

Answer

To solve the integral \( \int_{0}^{1} \frac{dx}{3x + 2} \), we will follow these steps:

### Step 1: Substitution
We start by making a substitution to simplify the integral. Let:
\[ z = 3x + 2 \]
Now, we need to find \( dx \) in terms of \( dz \). We differentiate both sides with respect to \( x \):
\[ \frac{dz}{dx} = 3 \implies dz = 3dx \implies dx = \frac{dz}{3} \]

### Step 2: Change the limits of integration
Next, we need to change the limits of integration according to our substitution. When \( x = 0 \):
\[ z = 3(0) + 2 = 2 \]
When \( x = 1 \):
\[ z = 3(1) + 2 = 5 \]
Thus, the new limits of integration are from \( z = 2 \) to \( z = 5 \).

### Step 3: Substitute in the integral
Now we substitute \( z \) and \( dx \) into the integral:
\[ \int_{0}^{1} \frac{dx}{3x + 2} = \int_{2}^{5} \frac{1}{z} \cdot \frac{dz}{3} \]
This simplifies to:
\[ \frac{1}{3} \int_{2}^{5} \frac{dz}{z} \]

### Step 4: Evaluate the integral
The integral \( \int \frac{dz}{z} \) is known to be \( \ln |z| \). Therefore, we have:
\[ \frac{1}{3} \left[ \ln |z| \right]_{2}^{5} = \frac{1}{3} \left( \ln(5) - \ln(2) \right) \]

### Step 5: Apply the properties of logarithms
Using the property of logarithms \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \):
\[ \frac{1}{3} \left( \ln(5) - \ln(2) \right) = \frac{1}{3} \ln\left(\frac{5}{2}\right) \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{1} \frac{dx}{3x + 2} \) is:
\[ \frac{1}{3} \ln\left(\frac{5}{2}\right) \]

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