Question:

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

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Partial fractions: $\frac{x}{(x+a)(x+b)} = \frac{1}{b-a} (\frac{b}{x+b} - \frac{a}{x+a})$.
Updated On: Apr 26, 2026
  • $\log \left( \frac{125}{16} \right)$
  • $\log \left( \frac{1024}{1125} \right)$
  • $\log \left( \frac{16}{125} \right)$
  • $\log \left( \frac{1125}{1024} \right)$
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The Correct Option is B

Solution and Explanation

Step 1: Partial Fractions
$\frac{x}{(x+2)(x+3)} = \frac{-2}{x+2} + \frac{3}{x+3}$.
Step 2: Integration
$[-2 \log(x+2) + 3 \log(x+3)]_1^2 = [\log \frac{(x+3)^3}{(x+2)^2}]_1^2$.
Step 3: Calculation
Value $= \log(\frac{5^3}{4^2}) - \log(\frac{4^3}{3^2}) = \log(\frac{125}{16}) - \log(\frac{64}{9}) = \log(\frac{125}{16} \times \frac{9}{64}) = \log(\frac{1125}{1024})$.
Wait, checking options... result is $\log(1125/1024)$.
Final Answer: (D)
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