# Prove that f(x)=2square rootx(lnx-2) is the antiderivative of f(x)=lnx/square rootx , x>0

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If (2*sqrt x)(ln x - 2) is the antiderivative of f(x) = (ln x) / (sqrt x), we have the derivative of (2*sqrt x)(ln x - 2) given by (ln x) / (sqrt x)

f(x) = (2*sqrt x)(ln x - 2)

f'(x) = [(2*sqrt x)(ln x - 2)]'

=> (2*sqrt x)'(ln x - 2) + (2*sqrt x)(ln x - 2)'

=> (2*(1/2)*x^(-1/2))(ln x - 2) + (2*sqrt x)(1/x)

=> (ln x - 2)/sqrt x + (2/sqrt x)

=> (1/sqrt x)*ln x

=> (ln x) / (sqrt x)

**This proves that (2*sqrt x)(ln x - 2) is the antiderivative of (ln x) / (sqrt x)**

To prove that the function F(x) = 2(sqrt x)(lnx-2) is the antiderivative of f(x) = lnx/sqrt x, we'll have to differentiate F(x).

F'(x) = f(x)

We'll differentiate F(x) using product rule:

F'(x) = 2(sqrt x)'*(lnx-2) + 2(sqrt x)*(lnx-2)'

F'(x) = (lnx-2)/sqrtx + (2sqrt x)/x

We'll multiply the 1st term by sqrtx:

F'(x) = (sqrtx)(lnx-2)/x + (2sqrt x)/x

F'(x) = (sqrtx)(lnx - 2 + 2)/x

F'(x) = (sqrtx)(lnx)/x

F'(x) = (lnx)/sqrt x = f(x)

**We notice that differentiating F(x), we've get f(x). Therefore F'(x) = f(x): [2(sqrt x)(lnx-2)]' = ln x/sqrt x.**