∫232√x−13−xdx
I tried to integrate at first with u=√x−3x−1 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=∫√x−13−xdx u=√x−13−x⟹x=3u2+1u2+1⟹dx=4u(u2+1)2du So I=∫4u2(u2+1)2dt=4∫duu2+1−4∫du(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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