Integration of x ln x

In this tutorial we shall find the integral of x ln x and solve this problem with the help of the integration by parts method.

The integral of x ln x is of the form

I = \int {x\ln xdx}

Here the first function is \ln x and the second function is x

I = \int {\ln x \cdot xdx} \,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Using the formula for integration by parts, we have

\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }

Using the formula above, equation (i) becomes

\begin{gathered} I = \ln x\int {xdx - \int {\left[ {\frac{d}{{dx}}\ln x\left( {\int {xdx} } \right)} \right]} dx} \\ \Rightarrow I = \ln x\frac{{{x^2}}}{2} - \int {\left[ {\frac{1}{x}\frac{{{x^2}}}{2}} \right]} dx \\ \Rightarrow I = \frac{{{x^2}}}{2}\ln x - \frac{1}{2}\int x dx \\ \Rightarrow I = \frac{{{x^2}}}{2}\ln x - \frac{1}{2}\frac{{{x^2}}}{2} + c \\ \Rightarrow I = \frac{{{x^2}}}{2}\ln x - \frac{1}{4}{x^2} + c \\ \end{gathered}