`int (dx)/(sqrt(4x^(2)-25))`

Answer

To solve the integral \( \int \frac{dx}{\sqrt{4x^2 - 25}} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{\sqrt{4x^2 - 25}} \]
Notice that \( 4x^2 - 25 \) can be rewritten as \( (2x)^2 - 5^2 \). This suggests that we can use a trigonometric substitution or a direct formula for integrals of the form \( \int \frac{dx}{\sqrt{x^2 - a^2}} \).

### Step 2: Substitute
Let \( t = 2x \). Then, we differentiate to find \( dx \):
\[ dt = 2dx \quad \Rightarrow \quad dx = \frac{1}{2} dt \]
Now we can rewrite the integral in terms of \( t \):
\[ I = \int \frac{\frac{1}{2} dt}{\sqrt{t^2 - 5^2}} = \frac{1}{2} \int \frac{dt}{\sqrt{t^2 - 25}} \]

### Step 3: Apply the Integral Formula
The integral \( \int \frac{dt}{\sqrt{t^2 - a^2}} \) has a standard result:
\[ \int \frac{dt}{\sqrt{t^2 - a^2}} = \ln |t + \sqrt{t^2 - a^2}| + C \]
In our case, \( a = 5 \). Thus, we have:
\[ I = \frac{1}{2} \left( \ln |t + \sqrt{t^2 - 25}| + C \right) \]

### Step 4: Substitute Back
Now we substitute back \( t = 2x \):
\[ I = \frac{1}{2} \left( \ln |2x + \sqrt{(2x)^2 - 25}| + C \right) \]
This simplifies to:
\[ I = \frac{1}{2} \ln |2x + \sqrt{4x^2 - 25}| + C \]

### Final Result
Thus, the integral \( \int \frac{dx}{\sqrt{4x^2 - 25}} \) is:
\[ \int \frac{dx}{\sqrt{4x^2 - 25}} = \frac{1}{2} \ln |2x + \sqrt{4x^2 - 25}| + C \]

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