<p>Whats the anti-derivative of ((lnx)^2)</p>
<p>Ahh, i don't think you have to know that for BC calc. But...</p>
<p>so would the final answer be 1/3(x(lnx)-x)^3</p>
<p>No, it's x(lnx)^2 - 2xlnx + 2x</p>
<p>I don't think you'd be given something this hard for AP calculus, but I just tried it and that's what I got. The way I did it was this (and if there's a simpler way, somebody please post... I just did this in 10 mins so I'm not sure if there's a simpler way):</p>
<p>edit: give me a minute, let me turn this into an image so it's easier to understand.</p>
<p>Here we go, this is how you do it:
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<p>Thanks, it was on my calc test</p>
<p>I got the same answer as GoldShadow, but with a MUCH easier method.</p>
<p>Basically, the problem is a integration by parts question within an integration by parts question. You setup normally, </p>
<p>u=lnx u'=1/x
v'=lnx v=∫lnx</p>
<p>Now for ∫(1)*lnx you do another setup another by parts equation:</p>
<p>u=lnx u'=1/x
v'=1 v=x</p>
<p>and you get xlnx-x as the result for ∫lnx</p>
<p>NOW going back to the original equation:</p>
<p>u=lnx u'=1/x
v'=lnx v=xlnx-x</p>
<p>which becomes:</p>
<p>lnx(xlnx-x)-∫lnx-1</p>
<p>And then you use the answer for ∫lnx YET another time to give:</p>
<p>lnx(xlnx-x)-(xlnx-x-x)</p>
<p>which simplifies to: x[(lnx)^2-2lnx+2]</p>
<p>This is a much simpler method (looks longer, but it is more logical...at least to me) and uses calculus you should know. Its a hard question; I'd be suprised to see this on an AP exam (hopefully not the one coming up for my sake :D)</p>