Integral Calculus Question 335
Question: $ \int_{{}}^{{}}{\frac{{{(x^{4}-x)}^{1/4}}}{x^{5}}\ dx} $ is equal to
Options:
A) $ \frac{4}{15}{{( 1-\frac{1}{x^{3}} )}^{5/4}}+c $
B) $ \frac{4}{5}{{( 1-\frac{1}{x^{3}} )}^{5/4}}+c $
C) $ \frac{4}{15}{{( 1+\frac{1}{x^{3}} )}^{5/4}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{{{(x^{4}-x)}^{1/4}}}{x^{5}},dx}=\int_{{}}^{{}}{\frac{1}{x^{4}}{{( 1-\frac{1}{x^{3}} )}^{1/4}}dx} $ $ =\frac{1}{3}\int_{{}}^{{}}{{t^{1/4}}dt}=\frac{4}{15}{t^{5/4}}+c=\frac{4}{15}{{( 1-\frac{1}{x^{3}} )}^{5/4}}+c $ $ { Putting,1-\frac{1}{x^{3}}=t,and,\frac{1}{x^{4}},dx=\frac{1}{3},dt } $
BETA
