Integral Calculus Question 395
Question: $ \int_{{}}^{{}}{\frac{1}{x\sqrt{1+\log x}}\ dx=} $
[Roorkee 1977]
Options:
A) $ \frac{2}{3}{{(1+\log x)}^{3/2}}+c $
B) $ {{(1+\log x)}^{3/2}}+c $
C) $ 2\sqrt{1+\log x}+c $
D) $ \sqrt{1+\log x}+c $
Show Answer
Answer:
Correct Answer: C
Solution:
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 $ .
BETA
