Integral Calculus Question 449
Question: $ \int_{{}}^{{}}{\frac{x^{2}{{\tan }^{-1}}x^{3}}{1+x^{6}}\ dx} $ is equal to
[MP PET 1999; UPSEAT 1999]
Options:
A) $ {{\tan }^{-1}}(x^{3})+c $
B) $ \frac{1}{6}{{({{\tan }^{-1}}x^{3})}^{2}}+c $
C) $ -\frac{1}{2}{{({{\tan }^{-1}}x^{3})}^{2}}+c $
D) $ \frac{1}{2}{{({{\tan }^{-1}}x^{2})}^{3}}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ x^{3}=t\Rightarrow dt=3x^{2},dx $ $ \int_{{}}^{{}}{\frac{x^{2}{{\tan }^{-1}}x^{3},dx}{1+x^{6}}}=\frac{1}{3}\int_{{}}^{{}}{\frac{{{\tan }^{-1}}t}{1+t^{2}}},dt $ Put $ z={{\tan }^{-1}}t\Rightarrow dz=\frac{dt}{1+t^{2}} $ $ =\frac{1}{3}\int_{{}}^{{}}{z,dz}=\frac{1}{3}\frac{z^{2}}{2}=\frac{z^{2}}{6}=\frac{1}{6}{{({{\tan }^{-1}}x^{3})}^{2}}+c $ .
BETA
