Question

Evaluate: \( \int \frac{dx}{(x+1)(x+2)} \).

Hint:

When using partial fractions, the "cover-up" method is a fast way to find coefficients for linear factors. To find A (the coefficient for \( \frac{1}{x+1} \)), cover up the \( (x+1) \) term in the original fraction and substitute \( x = -1 \) into the rest: \( A = \frac{1}{-1+2} = 1 \). Similarly for B, cover up \( (x+2) \) and substitute \( x = -2 \): \( B = \frac{1}{-2+1} = -1 \).

Answer 1

Step 1: Understanding the Concept:
The integral involves a rational function where the denominator is a product of linear factors. This suggests using the method of partial fraction decomposition to break down the integrand into simpler fractions that can be easily integrated.
Step 2: Key Formula or Approach:
We will decompose the fraction \( \frac{1}{(x+1)(x+2)} \) into the form: \[ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \] Multiplying both sides by \( (x+1)(x+2) \), we get: \[ 1 = A(x+2) + B(x+1) \] To find the constants A and B, we can substitute the roots of the denominator.
Step 3: Detailed Explanation:
First, let's find the values of A and B.
Set \( x = -1 \): \[ 1 = A(-1+2) + B(-1+1) \implies 1 = A(1) \implies A = 1 \] Next, set \( x = -2 \): \[ 1 = A(-2+2) + B(-2+1) \implies 1 = B(-1) \implies B = -1 \] Now, we can rewrite the integral as: \[ \int \frac{dx}{(x+1)(x+2)} = \int \left( \frac{1}{x+1} - \frac{1}{x+2} \right) dx \] Integrate each term separately: \[ \int \frac{1}{x+1} dx - \int \frac{1}{x+2} dx \] Using the standard integral \( \int \frac{1}{u} du = \ln|u| + C \): \[ = \ln|x+1| - \ln|x+2| + C \] Using the property of logarithms \( \ln a - \ln b = \ln(a/b) \): \[ = \ln \left| \frac{x+1}{x+2} \right| + C \] Step 4: Final Answer:
The evaluated integral is \( \ln \left| \frac{x+1}{x+2} \right| + C \).