`int(1)/(sqrt(4x+3))dx`

Answer

To solve the integral \( \int \frac{1}{\sqrt{4x + 3}} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( u = 4x + 3 \). Then, we differentiate to find \( dx \):
\[ \frac{du}{dx} = 4 \implies dx = \frac{du}{4} \]

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

### Step 3: Integrate
Now, we can integrate \( u^{-1/2} \):
\[ \int u^{-1/2} \, du = 2u^{1/2} + C \]
Thus, we have:
\[ \frac{1}{4} \cdot 2u^{1/2} + C = \frac{1}{2} u^{1/2} + C \]

### Step 4: Substitute Back
Now, substitute back \( u = 4x + 3 \):
\[ \frac{1}{2} \sqrt{4x + 3} + C \]

### Final Answer
The final result of the integral is:
\[ \int \frac{1}{\sqrt{4x + 3}} \, dx = \frac{1}{2} \sqrt{4x + 3} + C \]