Question

Solve $\int x^3\log xdx$

• A. $\frac{x^4\log x}{4}+C$
• B. $I=\frac{x^4}{4}\log x-\frac{x^4}{16}+C$
• C. $\frac{1}{8}\left[x^4\log x-4x^2\right]+C$
• D. $\frac{1}{16}\left[4x^4\log x+x^4\right]+C$

Answer 1

The correct option is B $I=\frac{x^4}{4}\log x-\frac{x^4}{16}+C$

Consider the given integral.

$I=\int x^3\log xdx$

$I=\log x\left(\frac{x^4}{4}\right)-\int \frac{1}{x}\left(\frac{x^4}{4}\right)dx$

$I=\frac{x^4}{4}\log x-\frac{1}{4}\int x^{3}dx$

$I=\frac{x^4}{4}\log x-\frac{1}{4}\left(\frac{x^4}{4}\right)+C$

$I=\frac{x^4}{4}\log x-\frac{x^4}{16}+C$

Hence, this is the answer.