Integral Calculus Question 13
Question: $ \int_{{}}^{{}}{\frac{2x{{\tan }^{-1}}x^{2}}{1+x^{4}}}\ dx= $
[Roorkee 1982]
Options:
A) $ {{[{{\tan }^{-1}}x^{2}]}^{2}}+c $
B) $ \frac{1}{2}{{[{{\tan }^{-1}}x^{2}]}^{2}}+c $
C) $ 2{{[{{\tan }^{-1}}x^{2}]}^{2}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ t={{\tan }^{-1}}x^{2}\Rightarrow dt=\frac{1}{1+x^{4}},2x,dx, $ then $ \int_{{}}^{{}}{\frac{2x{{\tan }^{-1}}x^{2}}{1+x^{4}}},dx=\int_{{}}^{{}}{t,dt}=\frac{t^{2}}{2}+c=\frac{1}{2}{{({{\tan }^{-1}}x^{2})}^{2}}+c. $
BETA
