MCQ
$\int_{}^{} {\frac{1}{x}\log x\;dx} $ is equal to
- A$\frac{1}{2}\log x + c$
- ✓$\frac{1}{2}{(\log x)^2} + c$
- C$\frac{1}{2}\log {(x)^2} + c$
- D$\log x + c$
✓
Answer
Correct option: B.
$\frac{1}{2}{(\log x)^2} + c$
b
(b) $I = \int_{}^{} {\frac{1}{x}\log x\,dx} $
(b) $I = \int_{}^{} {\frac{1}{x}\log x\,dx} $
Put $\log x = t \Rightarrow \frac{1}{x}\,dx = dt$
$\therefore \,\,\,I\int_{}^{} {t\,dt} = \frac{{{t^2}}}{2} + c = \frac{{{{(\log x)}^2}}}{2} + c$.
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