Integral Calculus Question 287
Question: $ \int_{{}}^{{}}{\frac{1}{x^{2}}{{(2x+1)}^{3}}}dx= $
Options:
A) $ 4x^{2}+12x+6\log x-\frac{1}{x}+c $
B) $ 4x^{2}+12x-6\log x-\frac{2}{x}+c $
C) $ 2x^{2}+8x+3\log x-\frac{2}{x}+c $
D) $ 8x^{2}+6x+6\log x+\frac{2}{x}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{1}{x^{2}}{{(2x+1)}^{3}}dx=\int_{{}}^{{}}{\frac{(8x^{3}+1+12x^{2}+6x)}{x^{2}},dx}} $ $ =\int_{{}}^{{}}{( 8x+12+\frac{6}{x}+\frac{1}{x^{2}} ),dx}=4x^{2}+12x+6\log x-\frac{1}{x}+c $ .
BETA
