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

Options

A. `(1)/(6)log|(x+5)/(x-1)|+c`
B. `(1)/(6)log|(x-1)/(x+5)|+c`
C. `(1)/(3)log|(x+5)/(x-1)|+c`
D. `(1)/(3)log|(x-1)/(x+5)|+c`

Answer

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

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

\[ x^2 + 4x - 5 \]

To factor this, we look for two numbers that multiply to \(-5\) (the constant term) and add to \(4\) (the coefficient of \(x\)). The numbers \(5\) and \(-1\) satisfy these conditions. Thus, we can factor the expression as:

\[ (x + 5)(x - 1) \]

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

\[ \int \frac{dx}{(x + 5)(x - 1)} \]

### Step 3: Use partial fraction decomposition
Next, we will use partial fraction decomposition to express the integrand as a sum of simpler fractions:

\[ \frac{1}{(x + 5)(x - 1)} = \frac{A}{x + 5} + \frac{B}{x - 1} \]

Multiplying through by the denominator \((x + 5)(x - 1)\) gives:

\[ 1 = A(x - 1) + B(x + 5) \]

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

\[ 1 = Ax - A + Bx + 5B \]

Combining like terms:

\[ 1 = (A + B)x + (5B - A) \]

Now, we can set up a system of equations by equating coefficients:

1. \(A + B = 0\)
2. \(5B - A = 1\)

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

\[ A = -B \]

Substituting into the second equation:

\[ 5B - (-B) = 1 \implies 5B + B = 1 \implies 6B = 1 \implies B = \frac{1}{6} \]

Now substituting back to find \(A\):

\[ A = -B = -\frac{1}{6} \]

### Step 5: Rewrite the integral with partial fractions
Now we can rewrite the integral:

\[ \int \left( \frac{-\frac{1}{6}}{x + 5} + \frac{\frac{1}{6}}{x - 1} \right) dx \]

This simplifies to:

\[ -\frac{1}{6} \int \frac{dx}{x + 5} + \frac{1}{6} \int \frac{dx}{x - 1} \]

### Step 6: Integrate each term
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, we can combine the logarithmic terms:

\[ \frac{1}{6} \left( \ln |x - 1| - \ln |x + 5| \right) = \frac{1}{6} \ln \left| \frac{x - 1}{x + 5} \right| + C \]

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

\[ \int \frac{dx}{x^2 + 4x - 5} = \frac{1}{6} \ln \left| \frac{x - 1}{x + 5} \right| + C \]