x^3 x , ,dx =

MCQ

$\int {{x^3}\log x\,\,dx = } $

A
$\frac{{{x^4}\log x}}{4} + c$
✓
$\frac{1}{{16}}[4{x^4}\log x - {x^4}] + c$
C
$\frac{1}{8}[{x^4}\log x - 4{x^2}] + c$
D
$\frac{1}{{16}}[4{x^4}\log x + {x^4}] + c$

✓

Answer

Correct option: B.

$\frac{1}{{16}}[4{x^4}\log x - {x^4}] + c$

b
(b) $I = \int {{x^3}\log x\,dx} $$ = \frac{{{x^4}}}{4}\log x - \int {\frac{{{x^4}}}{4}\frac{1}{x}dx + c} $
$ = \frac{{{x^4}}}{4}\log x - \int {\frac{{{x^3}}}{4}dx\, = \,\frac{{{x^4}}}{4}\log x - \frac{{{x^4}}}{{16}} + c} $
$ = \frac{1}{{16}}[4{x^4}\log x - {x^4}] + c$.

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