Saturday, May 11, 2019

calculus - How can I calculate the following integral (partial fractions)?



232x13xdx



I tried to integrate at first with u=x3x1 and after it I get a very complicated integration by partial fractions, can I solve it with an other way too? I think it has to be a much easier method, I appreciate any help.


Answer




You idea was very good.



ConsideringI=x13xdx u=x13xx=3u2+1u2+1dx=4u(u2+1)2du So I=4u2(u2+1)2dt=4duu2+14du(u2+1)2 The first one is simple; for the second one, use integration by parts.



I suppose that you did not work enough the simplification of dx.


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