<p>If I was doing it on a test though, Id be doing it with integration by parts, because that is what i learned, but both end up with the same answer in the end, so it doesn't really matter.</p>
<p>Wonky-Faint, I'm pretty sure my friend in AB is doing integration by parts now...</p>
<p>Yeah, parts is strictly BC material. If you're in doubt, check out the AP Calc course description on AP Central (<a href="http://apcentral.collegeboard.com/apc/members/repository/05836apcoursdesccalc0_4313.pdf)%5B/url%5D">http://apcentral.collegeboard.com/apc/members/repository/05836apcoursdesccalc0_4313.pdf)</a>. I'm pretty sure that link won't work if you're not logged in, but so be it.</p>
<p>Corroborator posted the correct AB method, but if you want to use parts, using the table method is probably easier than the formal way.</p>
<p>** u ** ///// ** dv ** ///// ** +/-**
x ///// (x-1)^1/2 ///// +
1 ///// (2/3)(x-1)^3/2 ///// -
0 ///// (4/15)(x-1)^5/2 ///// +
///////////////////////////////// -</p>
<p>so,</p>
<p>$ x(x-1)^1/2 dx = x(2/3)(x-1)^3/2 - (4/15)(x-1)^5/2</p>
<p>which is the same that christalena2 got.</p>