Question

How do you integrate `\int \sqrt { f ( x ) } f ^ { \prime } ( x ) d x`?

Answer 1

`intsqrt(f(x))f'(x)dx = 2/3f(x)sqrt(f(x))+"constant"`

Explanation:

let `F(x)-=f^((-1))(x) = intsqrt(f(x))f'(x)dx` and `u = f(x)` and `du = f'(x)dx`

`therefore F(x) = intsqrt udu = 2/3u^(3/2)+"constant" = 2/3[f(x)]^(3/2)+ "constant"`