\(\int\displaystyle\frac{1}{(x+1)(x+2)}\,dx=\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\log\displaystyle\frac{x+1}{x+2}+C\); where C is arbitrary constant of integration
By partial fractions, the two factors differ by 1, so
\( \displaystyle \frac{1}{(x+1)(x+2)} = \frac{1}{x+1}-\frac{1}{x+2} \)
(check: \((x+2)-(x+1) = 1\)). Integrating term by term:
\( \displaystyle \int\frac{dx}{(x+1)(x+2)} = \log|x+1|-\log|x+2|+C \)
\( \displaystyle = \log\left|\frac{x+1}{x+2}\right|+C \)
This is Option (2) (written without the modulus signs).
NCERT (2026-27 edition):
Integration by Partial Fractions
