`int(1)/(sqrt((x-2)(x-3)))dx=`

Options

A. `log|(x+(5)/(2))+sqrt(x^(2)+5x+6)|+c`
B. `log|(x-(5)/(2))+sqrt(x^(2)-5x+6)|+c`
C. `log|(x-(5)/(2))+sqrt(x^(2)+5x+6)|+c`
D. `log|(x-(5)/(2))+sqrt(x^(2)-5x-6)|+c`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1}{\sqrt{(x-2)(x-3)}} \, dx \]
We can expand the expression inside the square root:
\[ (x-2)(x-3) = x^2 - 5x + 6 \]
Thus, the integral becomes:
\[ \int \frac{1}{\sqrt{x^2 - 5x + 6}} \, dx \]

### Step 2: Complete the Square
Next, we complete the square for the quadratic expression \( x^2 - 5x + 6 \):
\[ x^2 - 5x + 6 = \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + 6 \]
Calculating the constant term:
\[ 6 = \frac{24}{4} \quad \text{so,} \quad -\frac{25}{4} + \frac{24}{4} = -\frac{1}{4} \]
Thus, we have:
\[ x^2 - 5x + 6 = \left(x - \frac{5}{2}\right)^2 - \frac{1}{4} \]
Now, substituting this back into the integral gives:
\[ \int \frac{1}{\sqrt{\left(x - \frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \, dx \]

### Step 3: Use the Standard Integral Formula
We can now use the standard result for the integral:
\[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln |x + \sqrt{x^2 - a^2}| + C \]
In our case, \( a = \frac{1}{2} \) and \( x \) is replaced with \( x - \frac{5}{2} \):
\[ \int \frac{1}{\sqrt{\left(x - \frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \, dx = \ln \left| x - \frac{5}{2} + \sqrt{\left(x - \frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2} \right| + C \]

### Step 4: Substitute Back
Substituting back, we have:
\[ \int \frac{1}{\sqrt{(x-2)(x-3)}} \, dx = \ln \left| x - \frac{5}{2} + \sqrt{\left(x - \frac{5}{2}\right)^2 - \frac{1}{4}} \right| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{\sqrt{(x-2)(x-3)}} \, dx = \ln \left| x - \frac{5}{2} + \sqrt{(x-2)(x-3)} \right| + C \]