Integral Calculus Question 321
Question: $ \int_{{}}^{{}}{\frac{{e^{5\log x}}-{e^{4\log x}}}{{e^{3\log x}}-{e^{2\log x}}}\ dx=} $
[MNR 1985]
Options:
A) $ e\ .\ {3^{-3x}}+c $
B) $ e^{3}\log x+c $
C) $ \frac{x^{3}}{3}+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{{e^{5\log x}}-{e^{4\log x}}}{{e^{3\log x}}-{e^{2\log x}}},dx}=\int_{{}}^{{}}{\frac{x^{5}-x^{4}}{x^{3}-x^{2}},dx} $ $ =\int_{{}}^{{}}{\frac{x^{4}(x-1)}{x^{2}(x-1)},dx}=\int_{{}}^{{}}{x^{2}dx}=\frac{x^{3}}{3}+c $ .
BETA
