Question

Prove the integration formula.

$\int \ln xdx=x\ln x-x+c$

(a) by applying integration by parts to $\int \ln xdx$.

(b) by differentiating$x\ln x-x$.

Short answer

(a) The solution is $\int \ln xdx=x\ln x-x+c$.

(b) The solution is$\ln x$.

Step-by-step solution

Part (a) Step 1. Given information.

It is given that$\int \ln xdx=x\ln x-x+c$.

Part (a) Step 2. Prove the given formula by applying the integration by parts.

Take $u=\ln x$and $v=1$.

$\begin{array}{rcl}\int \ln x·1dx& =& \ln x\int dx-\int \left[\frac{d}{dx}\left(\ln x\right)\int dx\right]dx\\ & =& x·\ln x-\int \left[\frac{1}{x}·x\right]dx\\ & =& x·\ln x-\int dx\\ & =& x·\ln x-x+C\end{array}$

Hence, the formula is proved.

Part (a) Step 1. Prove the formula by differentiating x lnx-x.

$\begin{array}{rcl}\frac{d}{dx}\left(x·\ln x-x\right)& =& \frac{d}{dx}\left(x·\ln x\right)-\frac{d}{dx}\left(x\right)\\ & =& x\frac{d}{dx}\left(\ln x\right)+\ln x\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(x\right)\\ & =& x·\frac{1}{x}+\ln x-1\\ & =& 1+\ln x-1\\ & =& \ln x\end{array}$

Hence, the formula is proved.