∫_1^2displaystyle1/(x(x+1)) dx, x gt 0 equals

The correct answer is Option (2) → \(\log\left(\displaystyle\frac{4}{3}\right)\)

Splitting into partial fractions:

    \( \displaystyle \frac{1}{x(x+1)} = \frac{(x + 1) - x}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1} \)

So

    \( \displaystyle \int_{1}^{2} \frac{1}{x(x+1)}\,dx = \Big[\log x - \log(x + 1)\Big]_{1}^{2} \)

    \( \displaystyle = \left[\log \frac{x}{x+1}\right]_{1}^{2} \)

    \( \displaystyle = \log \frac{2}{3} - \log \frac{1}{2} \)

    \( \displaystyle = \log \left(\frac{2}{3} \times 2\right) = \log \frac{4}{3} \)