Evaluate: `int1/(sqrt(4x^2-9))\ dx`

Answer

To evaluate the integral \( \int \frac{1}{\sqrt{4x^2 - 9}} \, dx \), we can follow these steps:

### Step 1: Rewrite the expression under the square root
We can express \( 4x^2 - 9 \) in a form that is easier to work with:
\[ 4x^2 - 9 = (2x)^2 - 3^2 \]
This allows us to recognize it as a difference of squares.

**Hint:** Look for a way to express the quadratic in a form that matches the standard integral formulas.

### Step 2: Identify the standard integral formula
We can use the formula:
\[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \log |x + \sqrt{x^2 - a^2}| + C \]
In our case, we have \( x = 2x \) and \( a = 3 \).

**Hint:** Familiarize yourself with standard integral formulas, especially those involving square roots.

### Step 3: Apply the formula
Using the formula, we substitute \( x = 2x \) and \( a = 3 \):
\[ \int \frac{1}{\sqrt{(2x)^2 - 3^2}} \, dx = \log |2x + \sqrt{(2x)^2 - 3^2}| + C \]
This simplifies to:
\[ \log |2x + \sqrt{4x^2 - 9}| + C \]

**Hint:** Make sure to substitute correctly and simplify the expression step by step.

### Step 4: Account for the derivative of the inner function
Since we have \( x = 2x \), we need to account for the derivative of \( 2x \) when applying the integral:
\[ \int \frac{1}{\sqrt{4x^2 - 9}} \, dx = \frac{1}{2} \log |2x + \sqrt{4x^2 - 9}| + C \]

**Hint:** Remember to adjust for the derivative of the inner function when using substitution.

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{1}{\sqrt{4x^2 - 9}} \, dx = \frac{1}{2} \log |2x + \sqrt{4x^2 - 9}| + C \]

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