Let ∫ (x1 / 2/√1-x3/2) d x=(2/3) operatornamegof(x)+C, then

Question

Let ∫ (x1 / 2/√1-x3/2) d x=(2/3) operatornamegof(x)+C, then (A) f ( x )=√ x  (B) f ( x )= x 3 / 2 and g ( x )= sin -1 x (C) f ( x )= x 2 / 3 (D) None of these

Tardigrade Admin · Accepted Answer

Put x 3 / 2= t ⇒ (3/2) x 1 / 2 dx = dt ∴ integral is ∫ ((2/3) dt /√1- t 2)=(2/3) sin -1 t + C =(2/3) sin -1( x 3 / 2)+ C

Q. Let $\int \frac{x ^{1/2}}{1 - x ^{3/2}} d x = \frac{2}{3} gof (x) + C$ , then

$f (x) = x$

$f (x) = x^{3/2}$ and $g (x) = sin^{- 1} x$

$f (x) = x^{2/3}$

None of these

Put $x^{3/2} = t \Rightarrow \frac{3}{2} x^{1/2} d x = d t$
$∴$ integral is
$\int \frac{\frac{2}{3} d t}{1 - t ^{2}} = \frac{2}{3} sin^{- 1} t + C$
$= \frac{2}{3} sin^{- 1} (x^{3/2}) + C$

Q. Let ∫1−x3/2​x1/2​dx=32​gof(x)+C, then

f(x)=x​

f(x)=x3/2 and g(x)=sin−1x

f(x)=x2/3