Integral Calculus Question 206
Question: $ \int_{{}}^{{}}{\frac{\log x\ dx}{x^{3}}=} $
[Roorkee 1986]
Options:
A) $ \frac{1}{4x^{2}}(2\log x-1)+c $
B) $ -\frac{1}{4x^{2}}(2\log x+1)+c $
C) $ \frac{1}{4x^{2}}(2\log x+1)+c $
D) $ \frac{1}{4x^{2}}(1-2\log x)+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{\log x}{x^{3}}dx=\int_{{}}^{{}}{{x^{-3}}\log x\ dx}} $ $ =-\frac{\log x}{2x^{2}}+\int_{{}}^{{}}{\frac{1}{x}.\frac{1}{2x^{2}}+c=-\frac{\log x}{2x^{2}}+\frac{1}{2}.\frac{{x^{-2}}}{-2}+c} $ $ =-\frac{\log x}{2x^{2}}-\frac{1}{4x^{2}}+c=-\frac{1}{4x^{2}}(2\log x+1)+c $ .
BETA
