• Sep 2nd 2008, 01:54 PM
andrewsx
integral ln/sqroot of x dx?
• Sep 2nd 2008, 02:05 PM
Chop Suey
Sub $u = \sqrt{x}$ and integrate by parts.
• Sep 2nd 2008, 02:06 PM
mr fantastic
Quote:

Originally Posted by andrewsx
integral ln/sqroot of x dx?

The obvious substitution is $u = \sqrt{x}$. Do it. Now use integration by parts.
• Sep 2nd 2008, 02:11 PM
o_O
$\int \ln \left(\sqrt{x}\right) dx$

You can use integration by parts: $\begin{array}{ll} u = \ln \left(\sqrt{x}\right) & dv = dx \\ du = \frac{1}{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \ dx = \frac{1}{2x} \ dx& v = x \end{array}$
• Sep 2nd 2008, 04:30 PM
Chop Suey
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. ;)