JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 8)
If $$\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx$$ = f(x) $$\sqrt {2x - 1} $$ + C, where C is a constant of integration, then f(x) is equal to :
$${2 \over 3}$$ (x $$-$$ 4)
$${1 \over 3}$$ (x + 4)
$${1 \over 3}$$ (x + 1)
$${2 \over 3}$$ (x + 2)
Explanation
$$\sqrt {2x - 1} = t \Rightarrow 2x - 1 = {t^2} \Rightarrow 2dx = 2t.dt$$
$$\int {{{x + 1} \over {\sqrt {2x - 1} }}dx = \int {{{{{{t^2} + 1} \over 2} + 1} \over t}tdt = \int {{{{t^2} + 3} \over 2}dt} } } $$
$$ = {1 \over 2}\left( {{{{t^3}} \over 3} + 3t} \right) = {t \over 6}\left( {{t^2} + 9} \right) + c$$
$$ = \sqrt {2x - 1} \left( {{{2x - 1 + 9} \over 6}} \right) + c = \sqrt {2x - 1} \left( {{{x + 4} \over 3}} \right) + c$$
$$ \Rightarrow f\left( x \right) = {{x + 4} \over 3}$$
$$\int {{{x + 1} \over {\sqrt {2x - 1} }}dx = \int {{{{{{t^2} + 1} \over 2} + 1} \over t}tdt = \int {{{{t^2} + 3} \over 2}dt} } } $$
$$ = {1 \over 2}\left( {{{{t^3}} \over 3} + 3t} \right) = {t \over 6}\left( {{t^2} + 9} \right) + c$$
$$ = \sqrt {2x - 1} \left( {{{2x - 1 + 9} \over 6}} \right) + c = \sqrt {2x - 1} \left( {{{x + 4} \over 3}} \right) + c$$
$$ \Rightarrow f\left( x \right) = {{x + 4} \over 3}$$
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