Integral Calculus Question 291
Question: $ \int_{{}}^{{}}{\frac{5(x^{6}+1)}{x^{2}+1}dx=} $
Options:
A) $ 5(x^{7}+x){{\tan }^{-1}}x+c $
B) $ x^{5}-\frac{5}{3}x^{3}+5x+c $
C) $ 3x^{4}-5x^{2}+15x+c $
D) $ 5{{\tan }^{-1}}(x^{2}+1)+\log (x^{2}+1)+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{5(x^{6}+1)}{x^{2}+1},dx=\int_{{}}^{{}}{\frac{5(x^{2}+1)(x^{4}-x^{2}+1)}{(x^{2}+1)},dx}} $ $ =\int_{{}}^{{}}{5(x^{4}-x^{2}+1),dx=x^{5}-\frac{5}{3}x^{3}+5x+c.} $
BETA
