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Calculus Level 3

What is the antiderivative

$\large \int \frac{\sqrt{x}}{\sqrt{x^3}}dx \ ?$

1

0

\frac{1}{x}

\ln(x) +c

1 solution

Alex Harman
Jun 5, 2016

First let's simplify $\large f'(x)=\frac{\sqrt{x}}{\sqrt{x^3}}=$
$\large x^{1/2} \times \frac{1}{x^{3/2}}=$ $\large x^{1/2} \times x^{-3/2}=x^{-1}$
Thus $\large f'(x)=x^{-1}$ ONLY when x>0 since x must be greater than 0 we have $\large \int x^{-1}dx=\ln(x)+c$ notice that $\large \ln\mid x\mid$ has values of x<0 and therefore f'(x) does not satisfy f(x)=ln|x|

Technically $\ln(|x| )+ c$ is also correct.

A Former Brilliant Member - 5 years ago

So what is the slope at $\f(-1)$ ? Input negative 1 into f'(x). And you get an imaginary value. It doesn't work.

Alex Harman - 5 years ago

The function basucally is
$f:(0,\infty) \to R$
$f(x) = \ln(|x|) + c$
I know that the modulus is pointless because of the domain, but that doesn't make it incorrect. Better just remove the option.

A Former Brilliant Member - 5 years ago

@A Former Brilliant Member – When f'(x) = sqrt x /sqrt x^3 it does NOT satisfy every value of ln |x|

Alex Harman - 5 years ago

@Alex Harman – Did you even read the domain I mentioned? I mentioned $(0,\infty)$ . The f'(x) satisfies all values in this range.

What I am saying is,
$x \in (0, \infty)$
$\ln(|x|) + c = \ln(x) + c$ thus making both answers correct.

I am not talking about negative values so don't bring that up.

A Former Brilliant Member - 5 years ago

@A Former Brilliant Member – The point is that the domain of f(x)=ln|x| is $(-\infty,\infty)$

Alex Harman - 5 years ago

@Alex Harman – We can choose to restrict the domain $(0,\infty)$

A Former Brilliant Member - 5 years ago

@A Former Brilliant Member – Okay I'll change it.

Alex Harman - 5 years ago

@A Former Brilliant Member – Not sure how to change the answers

Alex Harman - 5 years ago

@A Former Brilliant Member – $(-\infty,0)U(0,\infty)$ * sorry

Alex Harman - 5 years ago

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