Integral Calculus Question 30
Question: $ \int_{{}}^{{}}{\frac{dx}{(x^{2}+1)(x^{2}+4)}=} $
[MP PET 1995]
Options:
A) $ \frac{1}{3}{{\tan }^{-1}}(x)-\frac{1}{3}{{\tan }^{-1}}(\frac{x}{2})+c $
B) $ \frac{1}{3}{{\tan }^{-1}}(x)+\frac{1}{3}{{\tan }^{-1}}(\frac{x}{2})+c $
C) $ \frac{1}{3}{{\tan }^{-1}}(x)-\frac{1}{6}{{\tan }^{-1}}(\frac{x}{2})+c $
D) $ {{\tan }^{-1}}(x)-2{{\tan }^{-1}}(\frac{x}{2})+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{dx}{(x^{2}+1)(x^{2}+4)}}=\frac{1}{3}[ \int_{{}}^{{}}{\frac{dx}{x^{2}+1}-\int_{{}}^{{}}{\frac{dx}{x^{2}+4}}} ] $
$ =\frac{1}{3}[ {{\tan }^{-1}}x-\frac{1}{2}{{\tan }^{-1}}\frac{x}{2} ]+c=\frac{1}{3}{{\tan }^{-1}}x-(\frac{1}{6}{{\tan }^{-1}})(\frac{x}{2})+c $ .
BETA
