Answer
\[ = \frac{5}{9}{e^6} + \frac{1}{9}\]
Work Step by Step
\[\begin{gathered}
\int_1^{{e^2}} {{x^2}\ln x\,dx} \hfill \\
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set\,\,\,the\,\,substitution \hfill \\
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u = \ln x\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,du = \frac{1}{x}dx \hfill \\
dv = {x^2}\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,v = \frac{{{x^3}}}{3} \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
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{\text{replacing the values }}{\text{in the equation}} \hfill \\
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= \frac{{{x^3}\ln x}}{3} - \frac{1}{3}\int_{}^{} {\,\left( {\frac{{{x^3}}}{3}} \right)\,\left( {\frac{1}{x}} \right)\,dx} \hfill \\
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= \frac{{{x^3}\ln x}}{3} - \frac{1}{3}\int_{}^{} {{x^2}} dx \hfill \\
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integrate \hfill \\
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= \frac{{{x^3}\ln x}}{3} - \frac{1}{9}{x^3} + C \hfill \\
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therefore \hfill \\
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\,\,\int_1^{{e^2}} {{x^2}\ln x\,dx} = \left[ {\frac{{{x^3}\ln x}}{3} - \frac{1}{9}{x^3}} \right]_1^{{e^2}} \hfill \\
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evaluate\,\,the\,\,limits \hfill \\
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\,\left( {\frac{{{e^6}\ln {e^2}}}{3} - \frac{1}{9}{e^6}} \right) - \,\left( {0 - \frac{1}{9}} \right) \hfill \\
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{\text{Simplify}} \hfill \\
\hfill \\
= \frac{5}{9}{e^6} + \frac{1}{9} \hfill \\
\end{gathered} \]
