MCQ
The solution of $\frac{{dy}}{{dx}} = x\log x$ is
  • A
    $y = {x^2}\log x - \frac{{{x^2}}}{2} + c$
  • B
    $y = \frac{{{x^2}}}{2}\log x - {x^2} + c$
  • C
    $y = \frac{1}{2}{x^2} + \frac{1}{2}{x^2}\log x + c$
  • None of these

Answer

Correct option: D.
None of these
d
(d) $\frac{{dy}}{{dx}} = x\log x$ ==> $dy = x\log xdx$

==> $\int_{}^{} {dy = } \int_{}^{} {x\log xdx} $ ==>$y = \frac{{x^2}}{{2}} log\ x - \frac{{x^2}}{{4}} + c$.

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