\int {\Cos}^{-1} (\frac{1-x^2}{1+x^2}) \dx= [AP EAPCET 23 May 2025 S1]

Correct Answer is : 2[xTan−1x−log√1+x2]+c

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

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

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

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

AP EAPCET 23 May 2025 Shift 1 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

∫Cos−1(‌

1−x2

1+x2

)‌dx=

[AP EAPCET 23 May 2025 S1]

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