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. One may argue that we do not have two functions to apply this technique. That's true except that $\ln(x)$ may also be seen as the product of the functions $1$ and $\ln(x)$.

Set

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

Then

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

Since

\begin{displaymath}\int u dv = u v - \int v du\;,\end{displaymath}

we get

\begin{displaymath}\int x^n \ln(x) dx = \frac{x^{n+1}}{n+1} \ln(x) - \int \frac{x^{n+1}}{n+1} \frac{1}{x} dx \end{displaymath}

which implies

\begin{displaymath}\int x^n \ln(x) dx = \frac{x^{n+1}}{n+1} \ln(x) - \int \frac{x^n}{n+1}dx\end{displaymath}

or

\begin{displaymath}\int x^n \ln(x) dx = \frac{x^{n+1}}{n+1} \ln(x) - \frac{x^{n+1}}{(n+1)^2} + C\;.\end{displaymath}

It is a common mistake to forget the constant $C$.


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