Integral Calculus Question 490
Question: $ \int_{{}}^{{}}{\frac{{{({{\tan }^{-1}}x)}^{3}}}{1+x^{2}},dx=} $
[UPSEAT 2004]
Options:
A) $ {{({{\tan }^{-1}}x)}^{4}}+c $
B) $ \frac{{{({{\tan }^{-1}}x)}^{4}}}{4}+c $
C) $ 2{{\tan }^{-1}}x+c $
D) $ 2{{({{\tan }^{-1}}x)}^{2}}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ {{\tan }^{-1}}x=t\Rightarrow \frac{1}{1+x^{2}}dx=dt $ \ $ \int_{{}}^{{}}{\frac{{{({{\tan }^{-1}}x)}^{3}}}{1+x^{2}}}d\alpha =\int_{{}}^{{}}{t^{3}}dt=\frac{t^{4}}{4}+c $ = $ \frac{{{({{\tan }^{-1}}x)}^{4}}}{4}+c $ .
BETA
