Select all expressions matching the value of the definite integral int _(1)^e(x^2+1)/(x)dx (e^2+1)/(2) (e^2)/(2) i^2+1

To find the value of the definite integral $\int_{1}^{e} \frac{x^2 + 1}{x} \, dx$, we can simplify the integrand first:<br /><br />$\int_{1}^{e} \frac{x^2 + 1}{x} \, dx = \int_{1}^{e} (x + \frac{1}{x}) \, dx$<br /><br />Now, we can integrate term by term:<br /><br />$\int_{1}^{e} (x + \frac{1}{x}) \, dx = \int_{1}^{e} x \, dx + \int_{1}^{e} \frac{1}{x} \, dx$<br /><br />Integrating each term separately, we get:<br /><br />$\int_{1}^{e} x \, dx = \frac{x^2}{2} \bigg|_{1}^{e} = \frac{e^2}{2} - \frac{1^2}{2} = \frac{e^2}{2} - \frac{1}{2}$<br /><br />$\int_{1}^{e} \frac{1}{x} \, dx = \ln|x| \bigg|_{1}^{e} = \ln(e) - \ln(1) = 1 - 0 = 1$<br /><br />Adding the two results together, we get:<br /><br />$\int_{1}^{e} (x + \frac{1}{x}) \, dx = \frac{e^2}{2} - \frac{1}{2} + 1 = \frac{e^2}{2} + \frac{1}{2} = \frac{e^2 + 1}{2}$<br /><br />Therefore, the correct answer is $\frac{e^2 + 1}{2}$.