# Integrate the function$x\;logx$

$\begin{array}{1 1} \large \frac{\log x}{x^2}-\frac{x^2}{4}+c \\ \frac{\log x}{x^2}+\frac{x^2}{4}+c \\ \frac{\log x}{x}-\frac{x^2}{4}+c \\ \frac{\log x}{x}-\frac{x}{4}+c\end{array}$

Toolbox:
• (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts$\int udv=uv-\int vdu$
• (ii)$\int\frac{1}{x}dx=log x+c.$
Given $I=\int xlog xdx.$

Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts where $\int udv=uv-\int vdu$

Let u=log x.

On differentiating we get

$du=\frac{1}{x}dx.$

let dv=x dx.

On integrating we get

$v=\frac{x^2}{2}$.

On substituting for u,v,du and dv we get,

$\int xlog x=(log x.\frac{1}{x^2})-\int\frac{x^2}{2}.\frac{1}{x}dx.$

$\;\;\;\;=\frac{log x}{x^2}-\frac{1}{2}\int xdx.$

On integrating we get,

$\;\;\;\;=\frac{log x}{x^2}-\frac{1}{2}{x^2}{2}+c.$

$\;\;\;=\frac{log x}{x^2}-\frac{x^2}{4}+c.$

edited Feb 8, 2013