Question

The value of \( \int \frac{dx}{(x+1)(x+2)} \) is:

• A. \( \log \left| \frac{x-1}{x-2} \right| + c \)
• B. \( \log \left| \frac{x+2}{x+1} \right| + c \)
• C. \( \log \left| \frac{x+1}{x+2} \right| + c \)
• D. \( \log \left| \frac{x-1}{x+2} \right| + c \)

Hint:

For integration of rational functions, partial fractions can be a useful method to break the function into simpler fractions for easier integration.

Answer 1

The Correct Option is C

We are given the integral: \[ \int \frac{1}{(x+1)(x+2)} \, dx \] We will decompose the fraction into partial fractions: \[ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \] Multiplying both sides by \( (x+1)(x+2) \), we get: \[ 1 = A(x+2) + B(x+1) \] Expanding both sides: \[ 1 = A(x+2) + B(x+1) \] \[ 1 = A(x) + 2A + B(x) + B \] \[ 1 = (A + B)x + (2A + B) \] Now, comparing the coefficients of like terms: For the \( x \)-terms: \( A + B = 0 \) For the constant terms: \( 2A + B = 1 \) From \( A + B = 0 \), we have \( B = -A \). Substituting this into \( 2A + B = 1 \): \[ 2A - A = 1 \implies A = 1 \] Thus, \( B = -1 \). Therefore, we have the partial fraction decomposition: \[ \frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2} \] Now, integrating each term: \[ \int \frac{1}{x+1} \, dx = \log |x+1| \] \[ \int \frac{1}{x+2} \, dx = \log |x+2| \] Thus, the solution to the integral is: \[ \log |x+1| - \log |x+2| + c \] This can be written as: \[ \log \left| \frac{x+1}{x+2} \right| + c \] Thus, the correct answer is option (3).