`int(1/(x-3)-1/(x^(2)-3x))dx="________"+c, x gt 3`

Options

A. `1/3log[x(x-3)]`
B. `1/3log[sqrt(x)(x-3)]`
C. `2/3log[x(x-3)]`
D. `2/3log[sqrt(x)(x-3)]`

Answer

To solve the integral \(\int \left( \frac{1}{x-3} - \frac{1}{x^2 - 3x} \right) dx\), we can break it down into two separate integrals.

### Step 1: Rewrite the integral
The integral can be rewritten as:
\[ \int \left( \frac{1}{x-3} - \frac{1}{x(x-3)} \right) dx \]

### Step 2: Simplify the second term
The second term can be simplified:
\[ \frac{1}{x^2 - 3x} = \frac{1}{x(x-3)} \]
Thus, the integral becomes:
\[ \int \frac{1}{x-3} dx - \int \frac{1}{x(x-3)} dx \]

### Step 3: Integrate the first term
The first integral is straightforward:
\[ \int \frac{1}{x-3} dx = \ln |x-3| + C_1 \]

### Step 4: Integrate the second term using partial fractions
For the second integral, we can use partial fractions:
\[ \frac{1}{x(x-3)} = \frac{A}{x} + \frac{B}{x-3} \]
Multiplying through by \(x(x-3)\) gives:
\[ 1 = A(x-3) + Bx \]
Setting \(x = 0\) gives \(A(-3) = 1 \Rightarrow A = -\frac{1}{3}\).
Setting \(x = 3\) gives \(B(3) = 1 \Rightarrow B = \frac{1}{3}\).

Thus, we can rewrite:
\[ \frac{1}{x(x-3)} = -\frac{1}{3x} + \frac{1}{3(x-3)} \]

### Step 5: Integrate the second term
Now we can integrate:
\[ \int \left(-\frac{1}{3x} + \frac{1}{3(x-3)}\right) dx = -\frac{1}{3} \ln |x| + \frac{1}{3} \ln |x-3| + C_2 \]

### Step 6: Combine the results
Combining both integrals, we have:
\[ \int \left( \frac{1}{x-3} - \frac{1}{x^2 - 3x} \right) dx = \ln |x-3| - \left(-\frac{1}{3} \ln |x| + \frac{1}{3} \ln |x-3|\right) + C \]
This simplifies to:
\[ \ln |x-3| + \frac{1}{3} \ln |x-3| - \frac{1}{3} \ln |x| + C \]
\[ = \frac{4}{3} \ln |x-3| - \frac{1}{3} \ln |x| + C \]

### Final Result
Thus, the final answer is:
\[ \frac{4}{3} \ln |x-3| - \frac{1}{3} \ln |x| + C \]