Integral Calculus Question 153
Question: $ \int_{{}}^{{}}{{{\tan }^{-1}}xdx=} $
[Roorkee 1977]
Options:
A) $ x{{\tan }^{-1}}x+\frac{1}{2}\log (1+x^{2}) $
B) $ x{{\tan }^{-1}}x-\frac{1}{2}\log (1+x^{2}) $
C) $ (x-1){{\tan }^{-1}}x $
D) $ x{{\tan }^{-1}}x-\log (1+x^{2}) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{{{\tan }^{-1}}x dx}=x{{\tan }^{-1}}x-\int_{{}}^{{}}{\frac{x}{1+x^{2}} dx+c} $ $ =x{{\tan }^{-1}}x-\frac{1}{2}\log (1+x^{2})+c. $ Note : Students should remember this question as a formula.
BETA
