Integral Calculus Question 166
Question: $ \int_{{}}^{{}}{x{{\tan }^{-1}}}xdx= $
[Roorkee 1979]
Options:
A) $ \frac{1}{2}(x^{2}+1){{\tan }^{-1}}x-\frac{1}{2}x+c $
B) $ \frac{1}{2}(x^{2}-1){{\tan }^{-1}}x-\frac{1}{2}x+c $
C) $ \frac{1}{2}(x^{2}+1){{\tan }^{-1}}x+\frac{1}{2}x+c $
D) $ \frac{1}{2}(x^{2}+1){{\tan }^{-1}}x-x+c $
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Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{x,.,{{\tan }^{-1}}x,dx=\frac{x^{2}}{2}{{\tan }^{-1}}x-\frac{1}{2}\int_{{}}^{{}}{\frac{x^{2}+1-1}{1+x^{2}},dx}} $ $ =\frac{x^{2}}{2}{{\tan }^{-1}}x-\frac{1}{2}x+\frac{1}{2}{{\tan }^{-1}}x+c $ $ =\frac{1}{2}{{\tan }^{-1}}x,.,(x^{2}+1)-\frac{1}{2}x+c $ .
BETA
