\( \int_{1}^{2}\displaystyle\frac{dx}{(x+1)(x+2)}= \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \log_e 9-\log_e 8 \)
By partial fractions:
\( \displaystyle \frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2} \)
\( \displaystyle \int_{1}^{2}\frac{dx}{(x+1)(x+2)}=\Big[\log_e(x+1)-\log_e(x+2)\Big]_{1}^{2} \)
\( \displaystyle =\left[\log_e\frac{x+1}{x+2}\right]_{1}^{2} \)
\( \displaystyle =\log_e\frac{3}{4}-\log_e\frac{2}{3} \)
\( \displaystyle =\log_e\left(\frac{3}{4}\times\frac{3}{2}\right) \)
\( \displaystyle =\log_e\frac{9}{8} \)
\( \displaystyle =\log_e 9-\log_e 8 \)
Why the other options are incorrect
Option (1) \( \log_e 8-\log_e 9 \) — Incorrect. This is \( \log_e\frac{8}{9} \), which is negative; the integrand is positive on [1, 2], so the integral must be positive.
Option (3) \( \log_e 8+\log_e 9 \) — Incorrect. This is \( \log_e 72 \); a difference of logarithms has been turned into a sum.
Option (4) \( 2\log_e 9-\log_e 8 \) — Incorrect. This is \( \log_e\frac{81}{8} \), not \( \log_e\frac{9}{8} \).