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From: Pubkeybreaker on 17 Mar 2010 10:12 On Mar 16, 10:07 pm, Bart Goddard <goddar...(a)netscape.net> wrote: > Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> wrote innews:gerry-AC32AD..12543217032010(a)mail.eternal-september.org: > > > > > > > In article <Xns9D3DBBA686A78goddardbenetscape...(a)74.209.136.88>, > > Bart Goddard <goddar...(a)netscape.net> wrote: > > >> "J. Clarke" <jclarke.use...(a)cox.net> wrote in news:hnp2hq026a4 > >> @news3.newsguy.com: > > >> > On 3/16/2010 5:53 PM, Gerry Myerson wrote: > >> >> In article > >> >> <a6d89ea6-0081-4d28-be3b-772915b63...(a)l11g2000pro.googlegroups.com> > >> >> , > >> >> Leon<eduardoleonrodrig...(a)gmail.com> wrote: > > >> >>> integral (xe^(2x))/((2x+1)^2) BY PARTS > > >> >> Actually, I think it was by Newton, who preceded Parts > >> >> by many years. > > >> > Don't think so. Newton Senior's Parts preceded Newton by about 9 > >> > months. > > >> He's talking about "Issac Bertrand Parts", the pseudonym of Joseph > >> Quadratic, which he used when we was teaching at a > >> Jesuit school and wasn't allowed to publish under > >> his own name. > > > That was only one of his many pseudonyms. He also published > > as Jacob U. Substitution, Arnold Parshall Fractions, and > > Aloysius Change-The-Order-Of-Integration. > > Well, he sort of had to, after that scandal with > (pretty) Polly Nomial. > Multiply connected on his first intgration!
From: Robert H. Lewis on 17 Mar 2010 07:40 > 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. Robert H. Lewis Fordham University
From: Jay Belanger on 22 Mar 2010 23:20 "Robert H. Lewis" <rlewis(a)fordham.edu> writes: >> 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. Perhaps you've merely exposed the cutest joke of all.
From: Transfer Principle on 23 Mar 2010 19:03 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: uv - integral(vdu) = -xe^(2x)/(2(2x+1)) - integral(-e^(2x)(2x+1) / (2(2x+1)))dx = -xe^(2x)/(2(2x+1)) + 1/2 integral(e^(2x))dx = -xe^(2x)/(2(2x+1)) + e^(2x)/4 + C The LCD of these two fractions is 4(2x+1), so we have: = (-2xe^(2x) + e^(2x)(2x+1)) / (4(2x+1)) + C = (-2xe^(2x) + 2xe^(2x) + e^(2x)) / (4(2x+1)) + C = e^(2x)/((4(2x+1)) + C Thus there does exist an elementary antiderivative.
From: Robert H. Lewis on 23 Mar 2010 18:56
> 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! Robert H. Lewis Fordham University |