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

Options

A. `4log[(x-3)(x+1)]+c`
B. `(1)/(4)log[(x+1)/(x-3)]+c`
C. `(1)/(4)log[(x-3)/(x+1)]+c`
D. `(1)/(sqrt2)tan^(-1)[(x-1)/(sqrt2)]+c`

Answer

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

### Step 1: Factor the denominator
First, we need to factor the quadratic expression in the denominator:

\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]

### Step 2: Rewrite the integral
Now, we can rewrite the integral using the factored form:

\[ \int \frac{1}{(x - 3)(x + 1)} \, dx \]

### Step 3: Partial fraction decomposition
Next, we will express the integrand as a sum of partial fractions:

\[ \frac{1}{(x - 3)(x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1} \]

To find \(A\) and \(B\), we multiply through by the denominator \((x - 3)(x + 1)\):

\[ 1 = A(x + 1) + B(x - 3) \]

### Step 4: Solve for A and B
Expanding the right-hand side:

\[ 1 = Ax + A + Bx - 3B \]

Combining like terms gives:

\[ 1 = (A + B)x + (A - 3B) \]

Setting the coefficients equal, we have the system of equations:

1. \(A + B = 0\)
2. \(A - 3B = 1\)

From the first equation, we can express \(A\) in terms of \(B\):

\[ A = -B \]

Substituting into the second equation:

\[ -B - 3B = 1 \implies -4B = 1 \implies B = -\frac{1}{4} \]

Then substituting back to find \(A\):

\[ A = -(-\frac{1}{4}) = \frac{1}{4} \]

### Step 5: Rewrite the integral
Now we can rewrite the integral using the values of \(A\) and \(B\):

\[ \int \left( \frac{1/4}{x - 3} - \frac{1/4}{x + 1} \right) \, dx \]

### Step 6: Integrate term by term
Now we can integrate each term separately:

\[ \frac{1}{4} \int \frac{1}{x - 3} \, dx - \frac{1}{4} \int \frac{1}{x + 1} \, dx \]

The integrals are:

\[ \frac{1}{4} \ln |x - 3| - \frac{1}{4} \ln |x + 1| + C \]

### Step 7: Combine the logarithms
Using the properties of logarithms, we can combine the two logarithmic terms:

\[ \frac{1}{4} \left( \ln |x - 3| - \ln |x + 1| \right) = \frac{1}{4} \ln \left| \frac{x - 3}{x + 1} \right| + C \]

### Final Answer
Thus, the final result of the integral is:

\[ \int \frac{1}{x^2 - 2x - 3} \, dx = \frac{1}{4} \ln \left| \frac{x - 3}{x + 1} \right| + C \]

---