Integral Calculus Question 28
Question: $ \int_{{}}^{{}}{\frac{x^{2}}{(x^{2}+2)(x^{2}+3)}\ }dx= $
[AISSE 1990]
Options:
A) $ -\sqrt{2}{{\tan }^{-1}}(x)+\sqrt{3}{{\tan }^{-1}}(x)+c $
B) $ -\sqrt{2}{{\tan }^{-1}}(\frac{x}{\sqrt{2}})+\sqrt{3}{{\tan }^{-1}}(\frac{x}{\sqrt{3}})+c $
C) $ \sqrt{2}{{\tan }^{-1}}(\frac{x}{\sqrt{2}})+\sqrt{3}{{\tan }^{-1}}(\frac{x}{\sqrt{3}})+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{x^{2}}{(x^{2}+2)(x^{2}+3)}},dx=\int_{{}}^{{}}{[ \frac{3}{x^{2}+3}-\frac{2}{x^{2}+2} ]},dx $ $ =\frac{3}{\sqrt{3}}{{\tan }^{-1}}\frac{x}{\sqrt{3}}-\frac{2}{\sqrt{2}}{{\tan }^{-1}}( \frac{x}{\sqrt{2}} )+c $ $ =\sqrt{3}{{\tan }^{-1}}( \frac{x}{\sqrt{3}} )-\sqrt{2}{{\tan }^{-1}}( \frac{x}{\sqrt{2}} )+c. $
BETA
