Let $f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $. Then f(3)

Question

Let $f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $. Then f(3) – f(1) is eqaul to :

$ - {\pi \over {12}} + {1 \over 2} + {{\sqrt 3 } \over 4}$

$ {\pi \over {12}} + {1 \over 2} - {{\sqrt 3 } \over 4}$

$ - {\pi \over 6} + {1 \over 2} + {{\sqrt 3 } \over 4}$

${\pi \over 6} + {1 \over 2} - {{\sqrt 3 } \over 4}$

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