Answer
$\int_1^e\frac{(x^{2}+x+1)}{x}dx=\frac{e^{2}-1+2e}{2}$
Work Step by Step
Evaluate the integral $\int_1^e\frac{(x^{2}+x+1)}{x}dx$.
$\int_1^e(x+1+\frac{1}{x})dx$
$=\int_1^e(x)dx+\int_1^e(1)dx+\int_1^e(\frac{1}{x})dx$
$=[\frac{x^{2}}{2}]_1^e+[x]_1^e+ [lnx]_1^e$
$=\frac{1}{2}[e^{2}-1]+[e-1]+[lne-ln1]$
Hence, $\int_1^e\frac{(x^{2}+x+1)}{x}dx=\frac{e^{2}-1+2e}{2}$
