reduction formula (1 Viewer)

bleakarcher

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I(n)=integral[1/(x^2+1)^n] dx
find I(n) in terms of I(n-1)
 

Shadowdude

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Reduction formulae work via integration by parts - so.. I don't know. I may have a try at it, but I'm sure other maths geniuses will be able to spot the correct way sooner than I =P
 

muzeikchun852

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Reduction formulae work via integration by parts - so.. I don't know. I may have a try at it, but I'm sure other maths geniuses will be able to spot the correct way sooner than I =P
I think it's u=the whole thing / v'=1
 

Hermes1

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im pretty sure i have seen a question like this before and it usually comes with a preceding step that allows us to solve the question. OP is trolling us again.
 

Shadowdude

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Sorry for the double post but... are you sure that integral is correct?



?
 

bleakarcher

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deterministic
i got to that point but i dont know where to go after it. i could do it if it was in term of I(n+1), i would be able to do it
 

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