I thought it was interesting that u2+1(u2−2u−1)2 has the very simple integral −uu2−2u−1 but noth of u2(u2−wu−1)2 and 1(u2−2u−1)2 are very complicated (the transcendental parts cancel each other though). So my question is how do I check by looking at a rational function whether or not its

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I thought it was interesting that u2+1(u2−2u−1)2 has the very simple integral −uu2−2u−1 but noth of u2(u2−wu−1)2 and 1(u2−2u−1)2 are very complicated (the transcendental parts cancel each other though).  So my question is how do I check by looking at a rational function whether or not its

Marley Dupont · Accepted Answer

Examine the poles of your function (in the complex plane). If all residues are zero, you are in good shape.

besplodnexkj · Accepted Answer

To make a little more clear the reason why GEdgar&#039;s solution is works, observe that since all polynomials factor over C, you can take the rational function r and expand it completely using the method of partial fractions.That is, if r has poles at z1,z2,…,zk with multiplicities m1,m2,…,mk respectively, then for each iϵ{1,…,k}, and each jϵ{1,…,mi}, there is are constants aij such that the following holds.r(z)=a11z−z1+a12(z−z1)2+…+akmk(z−zk)mk.Every term here is integrable, except the terms ai1z−z1 where ai1≠0. This ai1 is exactly the residue of the function at zi. Therefore we get the criterion described by GEdgar.

I thought it was interesting that \frac{u^2+1}{(u^2-2u-1)^2} has the very simple integral -

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