MCQ
$\int_{}^{} {\frac{1}{{x\sqrt {1 + \log x} }}\;dx = } $
- A$\frac{2}{3}{(1 + \log x)^{3/2}} + c$
- B${(1 + \log x)^{3/2}} + c$
- ✓$2\sqrt {1 + \log x} + c$
- D$\sqrt {1 + \log x} + c$
✓
Answer
Correct option: C.
$2\sqrt {1 + \log x} + c$
c
(c)Put $t = 1 + \log x \Rightarrow dt = \frac{1}{x}dx$, then
$\int_{}^{} {\frac{{dx}}{{x\sqrt {1 + \log x} }}} = \int_{}^{} {\frac{{dt}}{{{t^{1/2}}}} = 2{t^{1/2}} + c} = 2{(1 + \log x)^{1/2}} + c$.
(c)Put $t = 1 + \log x \Rightarrow dt = \frac{1}{x}dx$, then
$\int_{}^{} {\frac{{dx}}{{x\sqrt {1 + \log x} }}} = \int_{}^{} {\frac{{dt}}{{{t^{1/2}}}} = 2{t^{1/2}} + c} = 2{(1 + \log x)^{1/2}} + c$.
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