Question

The value of the integral $\int_1^2 \frac{x \text{ d}x}{(x+2)(x+3)}$ is

• A. $\log \left( \frac{125}{16} \right)$
• B. $\log \left( \frac{1024}{1125} \right)$
• C. $\log \left( \frac{16}{125} \right)$
• D. $\log \left( \frac{1125}{1024} \right)$

Hint:

Partial fractions: $\frac{x}{(x+a)(x+b)} = \frac{1}{b-a} (\frac{b}{x+b} - \frac{a}{x+a})$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We resolve the rational function into partial fractions to perform the integration easily.
Step 2: Key Formula or Approach:
$\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}$
Using the cover-up method:
$A = \frac{-2}{-2+3} = -2$
$B = \frac{-3}{-3+2} = 3$
Step 3: Detailed Explanation:
The integral becomes:
\[ I = \int_1^2 \left( \frac{-2}{x+2} + \frac{3}{x+3} \right) \text{d}x \] \[ I = [-2 \log|x+2| + 3 \log|x+3|]_1^2 \] \[ I = (-2 \log 4 + 3 \log 5) - (-2 \log 3 + 3 \log 4) \] \[ I = -2 \log 4 + 3 \log 5 + 2 \log 3 - 3 \log 4 \] \[ I = 3 \log 5 + 2 \log 3 - 5 \log 4 \] \[ I = \log(5^3) + \log(3^2) - \log(4^5) = \log(125 \times 9) - \log(1024) \] \[ I = \log \left( \frac{1125}{1024} \right) \] Step 4: Final Answer:
The value of the integral is $\log \left( \frac{1125}{1024} \right)$.