`int(1)/(5+4x-x^(2))dx=`

Options

A. `(1)/(6)log[(x-1)/(x-5)]+c`
B. `(1)/(6)log[(x-5)/(x-1)]+c`
C. `(1)/(6)log[(1+x)/(x-5)]+c`
D. `(1)/(6)tan^(-1)[(x-5)/(x+1)]+c`

Answer

To solve the integral \( \int \frac{1}{5 + 4x - x^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the Denominator
First, we rewrite the expression in the denominator:
\[ 5 + 4x - x^2 = - (x^2 - 4x - 5) \]
This allows us to factor the quadratic expression:
\[ x^2 - 4x - 5 = (x - 5)(x + 1) \]
Thus, we have:
\[ 5 + 4x - x^2 = - (x - 5)(x + 1) \]

### Step 2: Set Up the Integral
Now, we can rewrite the integral:
\[ \int \frac{1}{5 + 4x - x^2} \, dx = -\int \frac{1}{(x - 5)(x + 1)} \, dx \]

### Step 3: Partial Fraction Decomposition
Next, we will use partial fraction decomposition to break down the integrand:
\[ \frac{1}{(x - 5)(x + 1)} = \frac{A}{x - 5} + \frac{B}{x + 1} \]
Multiplying through by the denominator \((x - 5)(x + 1)\) gives:
\[ 1 = A(x + 1) + B(x - 5) \]
Expanding this, we have:
\[ 1 = Ax + A + Bx - 5B = (A + B)x + (A - 5B) \]
Setting coefficients equal, we get the system:
1. \( A + B = 0 \)
2. \( A - 5B = 1 \)

### Step 4: Solve for A and B
From the first equation, we can express \( A \) in terms of \( B \):
\[ A = -B \]
Substituting into the second equation:
\[ -B - 5B = 1 \implies -6B = 1 \implies B = -\frac{1}{6} \]
Then substituting back to find \( A \):
\[ A = -(-\frac{1}{6}) = \frac{1}{6} \]

### Step 5: Rewrite the Integral
Now we can rewrite the integral:
\[ -\int \left( \frac{1/6}{x - 5} - \frac{1/6}{x + 1} \right) \, dx = -\frac{1}{6} \int \frac{1}{x - 5} \, dx + \frac{1}{6} \int \frac{1}{x + 1} \, dx \]

### Step 6: Integrate
Now we can integrate each term:
\[ -\frac{1}{6} \ln |x - 5| + \frac{1}{6} \ln |x + 1| + C \]

### Step 7: Combine the Logarithms
Using the properties of logarithms:
\[ \frac{1}{6} \left( \ln |x + 1| - \ln |x - 5| \right) = \frac{1}{6} \ln \left| \frac{x + 1}{x - 5} \right| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{1}{5 + 4x - x^2} \, dx = \frac{1}{6} \ln \left| \frac{x + 1}{x - 5} \right| + C \]