1. ## [SOLVED] integration

in this integration $\displaystyle \int \frac{lnx}{\sqrt{x}}dx$
is your $\displaystyle dv=x^\frac{1}{2}\$ or
$\displaystyle dv=x^\frac{-1}{2}\$

2. Hello,
Originally Posted by vinson24
in this integration $\displaystyle \int \frac{lnx}{\sqrt{x}}dx$
is your $\displaystyle dv=x^\frac{1}{2}\$ or
$\displaystyle dv=x^\frac{-1}{2}\$
I would recommend that you try both

Anyway, my reflex would be to take $\displaystyle dv=x^{\frac{1}{2}}$
because it will make appear a factor x while writing v : $\displaystyle v=\frac 23 \cdot x^{3/2}=\frac 23 \cdot x \cdot x^{1/2}$. And that x will simplify with du : $\displaystyle u=\ln(x) \Rightarrow du=\frac 1x$

But $\displaystyle dv=x^{-1/2}$ will work nicely too.

3. Originally Posted by vinson24
in this integration $\displaystyle \int \frac{lnx}{\sqrt{x}}dx$
is your $\displaystyle dv=x^\frac{1}{2}\$ or
$\displaystyle dv=x^\frac{-1}{2}\$
I will guess that your post means that you are trying to do the integration "by parts" (or perhaps that you are required to use "by parts").

What you use for u, v, du, and dv are entirely up to you. There can be more than one way to do a given integration. As long as your substitutions are valid and don't have you running in circles (ending up right back where you'd started) or heading in the wrong direction (getting only more complicated at each step), your choice is "right".

You can do the integration "by substitution", using $\displaystyle \sqrt{x}\, =\, u$, and you can always check your answer by re-differentiating and comparing with the original integrand (or else plugging the integrand into The Integrator on the MathWorld site).

For "by parts", the log is the messy bit, so I'd let ln(x) = u, so du = 1/x, dv = 1/sqrt{x} = x^(-1/2), and v = 2(x^(1/2)). That should get the job done.