Concept:Rewrite the integrand using the identity tan−1x+tan−1(1−x) to simplify the integral.Explanation:Let I=∫01cot−1(1+x2−x)dx.For 0≤x≤1, we have 1+x2−x>0, so cot−1(1+x2−x)=tan−1(1+x2−x1).Now, using the formula tan−1a+tan−1b=tan−1(1−aba+b) with a=x and b=1−x, we get tan−1x+tan−1(1−x)=tan−1(1−x(1−x)x+(1−x)).This is valid since x(1−x)≤41<1.Simplifying, tan−1x+tan−1(1−x)=tan−1(1−x+x21)=cot−1(1+x2−x).Thus, I=∫01[tan−1x+tan−1(1−x)]dx.Using the substitution t=1−x, we get ∫01tan−1(1−x)dx=∫01tan−1tdt.Therefore, I=2∫01tan−1xdx.Using integration by parts, ∫tan−1xdx=xtan−1x−21log(1+x2).Evaluating, ∫01tan−1xdx=[xtan−1x−21log(1+x2)]01=4π−21log2.So, I=2(4π−21log2)=2π−log2.Answer:2π−log2, which is Option A.