Question

The value of $\int x^3\log xdx$

• A. $\frac{1}{8}\left\{x^1\log x-4^4\right\}+C$
• B. $\frac{1}{16}\left\{4x\right\}$
• C. $\frac{x^4\log x}{4}C$
• D. None of these

Answer 1

The correct option is B $\frac{1}{16}\left\{4x\right\}$

Now,

$\int x^3\log xdx$

$=\log x.\frac{x^4}{4}-\int \frac{1}{x}.\frac{x^4}{4}dx$ [ Using method of by parts]

$=\frac{x^4\log x}{4}-\frac{x^4}{4^2}+c$ [Where is $c$ is integrating constant]