Answer
$\int_4^9(\sqrt x+{\frac{1}{\sqrt x}})^{2}dx=\frac{85}{2}+ln\frac{9}{4}$
Work Step by Step
Evaluate the integral $\int_4^9(\sqrt x+{\frac{1}{\sqrt x}})^{2}dx$.
$=\int_4^9(x+\frac{1}{x}+2)dx$
$=\int_4^9(x)dx+\int_4^9(\frac{1}{x})dx+2 \int_4^9(1)dx$
$=[\frac{x^{2}}{2}]_4^9+ln[x]_4^9+2 [x]_4^9$
$=\frac{1}{2}[9^{2}-4^{2}]+[ln9-ln4]+2 [9-4]$
Hence, $\int_4^9(\sqrt x+{\frac{1}{\sqrt x}})^{2}dx=\frac{85}{2}+ln\frac{9}{4}$