Question

$\int x\log \left(x\right)dx=?$

• A. $\frac{x^2}{4}[2\log (x)–1]+c$
• B. $\frac{x^2}{2}[2\log (x)–1]+c$
• C. $\frac{x^2}{4}[2\log (x)+1]+c$
• D. $\frac{x^2}{2}[2\log (x)+1]+c$

Answer 1

The correct option is A

$\frac{x^2}{4}[2\log (x)–1]+c$

Explanation For Correct The Option:

Evaluating the integral:

Given $\int x\log \left(x\right)dx$

Applying integration by parts,

$\int f\left(x\right).g\left(x\right).dx=f\left(x\right).\int g\left(x\right).dx-\int (f\text{'}(x).\int g(x).dx).dx$

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

Hence, the correct answer is option (A).