please help with this integral?

**andrewsx** · September 2nd 2008, 01:54 PM

integral ln/sqroot of x dx?

**Chop Suey** · September 2nd 2008, 02:05 PM

Sub $u = \sqrt{x}$ and integrate by parts.

**mr fantastic** · September 2nd 2008, 02:06 PM

Originally Posted by andrewsx

integral ln/sqroot of x dx?

The obvious substitution is $u = \sqrt{x}$ . Do it. Now use integration by parts.

**o_O** · September 2nd 2008, 02:11 PM

$\int \ln \left(\sqrt{x}\right) dx$

You can use integration by parts: $\begin{array}{ll} u = \ln \left(\sqrt{x}\right) & dv = dx \\ du = \frac{1}{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \ dx = \frac{1}{2x} \ dx& v = x \end{array}$

**Chop Suey** · September 2nd 2008, 04:30 PM

Alternatively, you can just rewrite the integral as:
$\frac{1}{2} \int \ln{x} dx$

I think it's obvious what u and dv should be in this case.

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