For 0< x< 1, \int [{\Tan}^{-1} (1-x+x^2) +{\Tan}^{-1}(1-x)] \dx= [AP EAPCET 24 May 2025 S1]

Correct Answer is : xCot−1x+log√1+x2+c

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

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

{\x} {\Cot}^{-1} {\x}+\frac{3}{4} \log (1+ {\x}^2) + {\c}

{\x} {\Tan}^{-1} {\x}-\frac{3}{4} \log \sqrt{1+ {\x}^2}+ {\c}

Indefinite Integration

Section: Mathematics

Question : 16 of 89

Marks: +1, -0

For 0<x<1,∫[Tan−1(1−x+x2)+Tan−1(1−x)]‌dx=

[AP EAPCET 24 May 2025 S1]

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