Integral Calculus Question 266
Question: $ \int_{{}}^{{}}{{{( x+\frac{1}{x} )}^{3}}}dx= $
Options:
A) $ \frac{1}{4}{{( x+\frac{1}{x} )}^{4}}+c $
B) $ \frac{x^{4}}{4}+\frac{3x^{2}}{2}+3\log x-\frac{1}{2x^{2}}+c $
C) $ \frac{x^{4}}{4}+\frac{3x^{2}}{2}+3\log x+\frac{1}{x^{2}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{{{( x+\frac{1}{x} )}^{3}}dx=\int_{{}}^{{}}{( x^{3}+\frac{1}{x^{3}}+3x+\frac{3}{x} ),dx}} $ $ =\frac{x^{4}}{4}-\frac{1}{2x^{2}}+\frac{3x^{2}}{2}+3\log x+c $ $ =\frac{x^{4}}{4}+\frac{3x^{2}}{2}+3\log x-\frac{1}{2x^{2}}+c. $
BETA
