JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 13)
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 :
$$ - {\pi \over {12}} + {1 \over 2} + {{\sqrt 3 } \over 4}$$
$$ {\pi \over {12}} + {1 \over 2} - {{\sqrt 3 } \over 4}$$
$$ - {\pi \over 6} + {1 \over 2} + {{\sqrt 3 } \over 4}$$
$${\pi \over 6} + {1 \over 2} - {{\sqrt 3 } \over 4}$$
Explanation
$$\int {{{\sqrt x } \over {{{(1 + x)}^2}}}} dx(x > 0)$$
Put x = tan2$$\theta $$ $$ \Rightarrow $$ 2xdx = 2tan$$\theta $$sec2$$\theta $$d$$\theta $$
$$I = \int {{{2{{\tan }^2}\theta .{{\sec }^2}\theta } \over 2}} d\theta = \int {2{{\sin }^2}\theta d\theta = \int {(1 - \cos 2\theta )d\theta } } $$
$$ = \theta - {{\sin 2\theta } \over 2} + c$$
$$ \Rightarrow f(x) = \theta - {1 \over 2} \times {{2\tan \theta } \over {1 + {{\tan }^2}\theta }} + c$$ $$f(x) = \theta - {{\tan \theta } \over {1 + {{\tan }^2}\theta }} + c = {\tan ^{ - 1}}\sqrt x - {{\sqrt x } \over {1 + x}} + c$$
Now $$f(3) - f(1) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {{\sqrt 3 } \over {1 + 3}} - {\tan ^{ - 1}}(1) + {1 \over 2}$$
$$ = {\pi \over 3} - {{\sqrt 3 } \over 4} - {\pi \over 4} + {1 \over 2}$$
$$ = {\pi \over 12} + {1 \over 2} - {{\sqrt 3 } \over 4}$$
Put x = tan2$$\theta $$ $$ \Rightarrow $$ 2xdx = 2tan$$\theta $$sec2$$\theta $$d$$\theta $$
$$I = \int {{{2{{\tan }^2}\theta .{{\sec }^2}\theta } \over 2}} d\theta = \int {2{{\sin }^2}\theta d\theta = \int {(1 - \cos 2\theta )d\theta } } $$
$$ = \theta - {{\sin 2\theta } \over 2} + c$$
$$ \Rightarrow f(x) = \theta - {1 \over 2} \times {{2\tan \theta } \over {1 + {{\tan }^2}\theta }} + c$$ $$f(x) = \theta - {{\tan \theta } \over {1 + {{\tan }^2}\theta }} + c = {\tan ^{ - 1}}\sqrt x - {{\sqrt x } \over {1 + x}} + c$$
Now $$f(3) - f(1) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {{\sqrt 3 } \over {1 + 3}} - {\tan ^{ - 1}}(1) + {1 \over 2}$$
$$ = {\pi \over 3} - {{\sqrt 3 } \over 4} - {\pi \over 4} + {1 \over 2}$$
$$ = {\pi \over 12} + {1 \over 2} - {{\sqrt 3 } \over 4}$$
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