Answer
$\displaystyle \frac{7}{3}$
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Work Step by Step
Find the indefinite integral first,
$\displaystyle \int\frac{(1+\ln x)^{2}}{x}dx=\left[\begin{array}{l}
u=1+\ln x\\
du=\frac{1}{x}dx
\end{array}\right]$
$=\displaystyle \int u^{2}du=\frac{u^{3}}{3}+C$
$=\displaystyle \frac{1}{3}(1+\ln x)^{3}+C$
Now, the definite integral:
$\displaystyle \int_{1}^{e}\frac{(1+\ln x)^{2}}{x}dx=[\frac{1}{3}(1+\ln x)^{3}]_{1}^{e}$
$=\displaystyle \frac{1}{3}[1+\ln e)^{3}-(1+\ln 1)^{3}]$
$= \displaystyle \frac{1}{3}(8-1) =\frac{7}{3}$
