JEE Main 20204 Sep, Morning shiftMathematicsMediumIntegral Calculus369× since 2002Let f(x)=∫x(1+x)2dx(x≥0)f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)}f(x)=∫(1+x)2xdx(x≥0). Then f(3) – f(1) is eqaul to :AOption A. −π12+12+34- {\pi \over {12}} + {1 \over 2} + {{\sqrt 3 } \over 4}−12π+21+43BOption B. π12+12−34{\pi \over {12}} + {1 \over 2} - {{\sqrt 3 } \over 4}12π+21−43COption C. −π6+12+34- {\pi \over 6} + {1 \over 2} + {{\sqrt 3 } \over 4}−6π+21+43DOption D. π6+12−34{\pi \over 6} + {1 \over 2} - {{\sqrt 3 } \over 4}6π+21−43CheckSkip · show solution
