The value of \(\int_{1}^{2}\displaystyle\frac{dx}{x(x+1)}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\log_e\displaystyle\frac{4}{3}\)
By partial fractions,
\( \displaystyle \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1} \)
So
\( \displaystyle \int_{1}^{2}\frac{dx}{x(x+1)} = \left[\log_e x - \log_e(x+1)\right]_{1}^{2} \)
\( = (\log_e 2 - \log_e 3) - (0 - \log_e 2)\)
\( \displaystyle = 2\log_e 2 - \log_e 3 = \log_e\frac{4}{3} \)