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From: Robert Israel on 24 Mar 2010 16:08 > > On Mar 17, 8:40 am, "Robert H. Lewis" > > <rle...(a)fordham.edu> wrote: > > > > integral (xe^(2x))/((2x+1)^2) BY PARTS > > > Hate to break into all the cute jokes, but the OP's problem > > > probably has a typo, as it cannot be expressed with elementary > > > functions. > > > > I disagree. I believe that it can be so expressed. > > Since the OP first wrote the problem a week ago and most teachers > > don't give more than a week to complete the assignment, I feel > > that I can post this without spoiling the homework. > > > > Let u = xe^(2x) and dv = dx/((2x+1)^2). Then v = > > -1/(2(2x+1) and > > du = (x(e^(2x))2 + e^(2x))dx = e^(2x)(2x+1)dx. Hey, > > that's convenient -- du has a factor of (2x+1) in the > > numerator and v has a factor of (2x+1) in the denominator! > > > > So integrating by parts, we have:... > > = e^(2x)/((4(2x+1)) + C > > > > Thus there does exist an elementary antiderivative. > > You're right. A week ago when I posted that I just lazily fed the integral > to Maple12. It responded: > > -(1/2)*ln(xe)*xe^(2*x)/(2*ln(xe)*x+ln(xe))-(1/2)*ln(xe)*Ei(1, > -2*ln(xe)*x-ln(xe))/xe > > with the Ei (exponential integral) function. Knowing a bit about Maple > (not too much) I applied simplify to the above and got > > -(1/2)*(xe^(2*x+1)+2*ln(xe)*Ei(1, -ln(xe)*(2*x+1))*x+ln(xe)*Ei(1, > -ln(xe)*(2*x+1)))/((2*x+1)*xe) > > at which point I trusted Maple and posted the above. Shame on me for > trusting Maple! A little knowledge is a dangerous thing. It seems you fed Maple int(xe^(2*x)/(2*x+1)^2, x); and Maple interprets xe as a new variable. You should have tried int(x*exp(2*x)/(2*x+1)^2, x); which would return the elementary antiderivative 1/4 * exp(2*x)/(2*x+1) -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada |