Integral Calculus Question 183
Question: $ \int_{{}}^{{}}{\frac{3x^{2}}{x^{6}+1}dx=} $
[MNR 1981; MP PET 1988; RPET 1995]
Options:
A) $ \log (x^{6}+1)+c $
B) $ {{\tan }^{-1}}(x^{3})+c $
C) $ 3{{\tan }^{-1}}(x^{3})+c $
D) $ 3{{\tan }^{-1}}( \frac{x^{3}}{3} )+c $
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Answer:
Correct Answer: B
Solution:
Put $ x^{3}=t\Rightarrow 3x^{2}dx=dt, $ therefore $ \int_{{}}^{{}}{\frac{3x^{2}}{x^{6}+1},dx=\int_{{}}^{{}}{\frac{1}{t^{2}+1}dt={{\tan }^{-1}}(t)+c}}={{\tan }^{-1}}(x^{3})+c $ .
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