The value of \(\int_{1}^{2}\displaystyle\frac{x+1}{x^{2}}\,dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\log_e 2+\displaystyle\frac{1}{2}\)
Dividing each term of the numerator by \(x^{2}\):
\( \displaystyle \frac{x+1}{x^{2}} = \frac{1}{x} + \frac{1}{x^{2}} \)
So
\( \displaystyle \int_{1}^{2}\left(\frac{1}{x} + \frac{1}{x^{2}}\right)dx = \left[\log_e x - \frac{1}{x}\right]_{1}^{2} \)
\( \displaystyle = \left(\log_e 2 - \frac{1}{2}\right) - \left(0 - 1\right) \)
\( \displaystyle = \log_e 2 + \frac{1}{2} \)