Problems on Techniques of Integration

The logarithmic function is one of these functions whose derivative is very nice (rational function). So it is a good idea to use the integration by parts technique in which $\ln(x)$ will be differentiated. There are basically two ways of doing this problem. You may try to find the antiderivative of $x^2 \ln(x)$ then takes its values at $e$ and $1$. Or you may want to use the integration by parts for definite integrals.

Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=& \ln(x)\\
dv &=& x^2 dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&\displaystyle \frac{1}{x} dx\\
&&\\
v &=& \displaystyle \frac{x^3}{3}\;.
\end{array}\right.\end{displaymath}

Since

\begin{displaymath}\int_a^b u dv = [u v]_a^b - \int_a^b v du\;,\end{displaymath}

we get

\begin{displaymath}\int_1^e x^2 \ln(x) dx = \left[\frac{x^3}{3} \ln(x)\right]_1^...
...left[\frac{x^3}{3} \ln(x)\right]_1^e - \int_1^e \frac{x^2}{3}dx\end{displaymath}

which implies

\begin{displaymath}\int_1^e x^2 \ln(x) dx = \left[\frac{x^3}{3} \ln(x)\right]_1^...
..._1^e = \frac{e^3}{3} - \left(\frac{e^3}{9} - \frac{1}{9}\right)\end{displaymath}

or

\begin{displaymath}\int_1^e x^2 \ln(x) dx = \frac{(2e^3 + 1)}{9}\cdot\end{displaymath}


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