JEE Main 201910 Apr, Morning shiftMathematicsMediumIntegral Calculus369× since 2002If ∫dx(x2−2x+10)2=A(tan−1(x−13)+f(x)x2−2x+10)+C\int {{{dx} \over {{{\left( {{x^2} - 2x + 10} \right)}^2}}}} = A\left( {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{f\left( x \right)} \over {{x^2} - 2x + 10}}} \right) + C∫(x2−2x+10)2dx=A(tan−1(3x−1)+x2−2x+10f(x))+C where C is a constant of integration then :AOption A. A =154{1 \over {54}}541 and f(x) = 9(x–1)²BOption B. A =154{1 \over {54}}541 and f(x) = 3(x–1)COption C. A =181{1 \over {81}}811 and f(x) = 3(x–1)DOption D. A =127{1 \over {27}}271 and f(x) = 9(x–1)²CheckSkip · show solution
