MHT CET · Maths · Definite Integration
The value of the integral \(\int_1^2 \frac{x \mathrm{~d} x}{(x+2)(x+3)}\) is
- A \(\log \left(\frac{125}{16}\right)\)
- B \(\log \left(\frac{1024}{1125}\right)\)
- C \(\log \left(\frac{16}{125}\right)\)
- D \(\log \left(\frac{1125}{1024}\right)\)
Answer & Solution
Correct Answer
(D) \(\log \left(\frac{1125}{1024}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{x}{(x+2)(x+3)} = \frac{3}{x+3} - \frac{2}{x+2}\) \(\int_1^2 \left(\frac{3}{x+3} - \frac{2}{x+2}\right) \mathrm{d} x = \left[\ln\left(\frac{(x+3)^3}{(x+2)^2}\right)\right]_1^2\)
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