Definite Integration Question 485
Question: The value of $ \int_0^{1}{\frac{x^{4}+1}{x^{2}+1}dx} $ is
[MP PET 1998]
Options:
A) $ \frac{1}{6}(3\pi -4) $
B) $ \frac{1}{6}(3-4\pi ) $
C) $ \frac{1}{6}(3\pi +4) $
D) $ \frac{1}{6}(3+4\pi ) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_0^{1}{\frac{x^{4}+1}{x^{2}+1}dx=\int_0^{1}{\frac{x^{4}-1}{x^{2}+1}dx+2\int_0^{1}{\frac{dx}{1+x^{2}}}}} $
therefore $ I=\int_0^{1}{(x^{2}-1)}dx+2\int_0^{1}{\frac{dx}{1+x^{2}}} $
therefore $ I=[ \frac{x^{3}}{3}-x ]_0^{1}+2[{{\tan }^{-1}}x]_0^{1} $
$ =-\frac{2}{3}+\frac{\pi }{2}=\frac{(3\pi -4)}{6} $ .
BETA
