Integrate the function$x\;log2x$

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

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 $I=\int xlog 2xdx.$

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 2x.

On differentiating with respect to x,

$du=2\frac{1}{2x}dx=\frac{1}{x}dx.$

let dv=x dx.

On integrating we get,

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

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

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

On cancelling x we get

$I=\frac{x^2log 2x}{2}-\frac{1}{2}\int x dx.$

On integrating we get,

$\int xlog 2xdx=\frac{x^2log 2x}{2}-\frac{x^2}{4}+c.$

edited Feb 8, 2013