A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
[b] \[\int_{0}^{1}{{{\cot }^{-1}}(1-x+{{x}^{2}})dx}\] \[=\int_{0}^{1}{{{\tan }^{-1}}\left( \frac{1}{1-x+{{x}^{2}}} \right)}dx\] \[=\int_{0}^{1}{{{\tan }^{-1}}\left( \frac{x+(1-x)}{1-x(1-x)} \right)}dx\] \[=\int_{0}^{1}{{{\tan }^{-1}}xdx+\int_{0}^{1}{{{\tan }^{-1}}(1-x)dx}}\] \[=\int_{0}^{1}{{{\tan }^{-1}}xdx+\int_{0}^{1}{{{\tan }^{-1}}[1-(1-x)]dx}}\] \[=2\int_{0}^{1}{{{\tan }^{-1}}xdx}\]or \[\lambda =2\]
You need to login to perform this action.
You will be redirected in
3 sec
