Integral Calculus Question 357
Question: $ \int_{{}}^{{}}{\frac{1}{{{[{{(x-1)}^{3}}{{(x+2)}^{5}}]}^{1/4}}}\ dx} $ is equal to
Options:
A) $ \frac{4}{3}{{( \frac{x-1}{x+2} )}^{1/4}}+c $
B) $ \frac{4}{3}{{( \frac{x+2}{x-1} )}^{1/4}}+c $
C) $ \frac{1}{3}{{( \frac{x-1}{x+2} )}^{1/4}}+c $
D) $ \frac{1}{3}{{( \frac{x+2}{x-1} )}^{1/4}}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{1}{{{[{{(x-1)}^{3}}{{(x+2)}^{5}}]}^{1/4}}}},dx=\int_{{}}^{{}}{\frac{1}{{{( \frac{x-1}{x+2} )}^{3/4}}{{(x+2)}^{2}}}},dx $ $ =\frac{1}{3}\int_{{}}^{{}}{\frac{1}{{t^{3/4}}},dt} $ , $ { \because ,\frac{x-1}{x+2}=t\Rightarrow \frac{3}{{{(x+2)}^{2}}},dx=dt } $ $ =\frac{1}{3}( \frac{{t^{1/4}}}{1/4} )+c=\frac{4}{3}{t^{1/4}}+c=\frac{4}{3}{{( \frac{x-1}{x+2} )}^{1/4}}+c $ .
BETA
