When x>0, then \int {\Cos}^{-1} (\frac{1-x^2}{1+x^2}) \dx is ........................

Correct Answer is : 2xTan−1x−log(1+x2)+c

2 [x {\Tan}^{-1} x-\log (1+x^2) ] +c

2 [x {\Tan}^{-1} x+\log (1+x^2) ] +c

2 x {\Tan}^{-1} x+\log (1+x^2) +c

2 x {\Tan}^{-1} x-\log (1+x^2) +c

KCET 2011 Math Solved Paper

Question : 52 of 60

Marks: +1, -0

When x>0, then ∫Cos−1(‌

1−x2

1+x2

)‌dx is ........................

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