Integral Calculus Question 32
Question: $ \int{\frac{x^{2}}{x^{2}+4}dx} $ equals to
[RPET 2001]
Options:
A) $ x-2{{\tan }^{-1}}(x/2)+c $
B) $ x+2{{\tan }^{-1}}(x/2)+c $
C) $ x-4{{\tan }^{-1}}(x/2)+c $
D) $ x+4{{\tan }^{-1}}(x/2)+c $
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Answer:
Correct Answer: A
Solution:
$ I=\int{\frac{x^{2}}{x^{2}+4}dx} $ $ =\int{\frac{x^{2}+4-4}{(x^{2}+4)}dx} $
Þ $ I=\int{[ 1-\frac{4}{x^{2}+4} ],dx} $ $ =\int{dx-\int{\frac{4}{x^{2}+4}dx}} $
$ \Rightarrow I=x-4\int{\frac{dx}{x^{2}+{{(2)}^{2}}}} $ $ =x-\frac{4}{2}{{\tan }^{-1}}(x/2)+c $ $ =x-2{{\tan }^{-1}}\frac{x}{2}+c $ .
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