Question

$\int_{}^{}\frac{\log(x + 1)–\log x}{x(x + 1)}$ dx is equal to –

• A. $–\frac{1}{2}\left\lbrack \log\left( \frac{x + 1}{x} \right) \right\rbrack^{2} + C$
• B. C – [{log (x + 1)}² – (log x)²]
• C. $–\log\left\lbrack \log\left( \frac{x + 1}{x} \right) \right\rbrack + C$
• D. – log $\left( \frac{x + 1}{x} \right)$+ C

Answer 1

Answer: (A)

$–\frac{1}{2}\left\lbrack \log\left( \frac{x + 1}{x} \right) \right\rbrack^{2} + C$

Answer 2

Put log (x + 1) – log x = t

̃ ̃ $\frac{x–(x + 1)}{x(x + 1)} = \frac{dt}{dx}$

̃ = dt̃ $\frac{dx}{x(x + 1)}$ = – dt

so question becomes

– $\int$t dt = – $\frac{t^{2}}{2} + C$