Question

What is the integration of $\ln x$?

Answer 1

Find the integration of $\ln x$.

From Integration by parts $\begin{array}{rcl}\mathbf{\int }\mathit{u}\mathit{v}\mathbf{}\mathit{d}\mathit{x}& \mathbf{=}& \mathbf{u}\mathbf{\int }\mathbf{v}\mathbf{}\mathbf{d}\mathbf{x}\mathbf{-}\mathbf{\int }\left(\int v\mathrm{dx}\right)\mathbf{\left(}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{\right)}\mathbf{}\mathbf{d}\mathbf{x}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\end{array}$

Put $u=\ln \left(\mathrm{x}\right)andv=1$ in the equation$\left(1\right)$:

$\begin{array}{rcl}\int \ln xdx& =& \int \ln x.1dx\\ & \Rightarrow & \int \ln xdx=\left[\ln x.\left(x\right)-\int x\left(\frac{1}{x}\right)dx\right]+C\left[\because C\mathrm{is}\mathrm{a}\mathrm{constant}.\right]\\ & \Rightarrow & \int \ln xdx=x\left(\ln x\right)-x+C\end{array}$

Hence, the required integral is $\int \ln x=x\left(\ln x\right)-x+C$ ,where $C$ is a constant.