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I do not think so. I gave it some thought but nothing springs to mind. That is one seriously nasty integral though ¬___________¬
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I know.
I just hope they aren't gonna give questions like that on the test :/ |
I've been doing some more thinking on your problem and have come up with a possible simplification.
You could try the substitution t = x^k^-1, where k is any real number. What that does is inverse the power so you still end up with t^5 and t^4 on the denominator but it should simplify the numerator and make the integrand easier to tackle. I haven't run through the maths or anything but if you want, try it and see. |
You mean take t = x^(1/20)? (t = x^20^-1 ?)
I tried that, gives a long answer. It requires applying the binomial theorem with a power of 15) Or something else like t = 1/x^k? How? :/ |
I meant what I typed out exactly.
t = x^k^-1 So for the first term you have x^(1/5) so the transformation gives t^5 and the second term goes from x^(1/4) to t^4 If that makes sense. Edit: The transformation inverses the power of x or in other words you flip the fractions over. |
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