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anonymous
 5 years ago
Integration: Out of the following methods, substitution/ parts/ partial fractions, which methods would you choose to solve each of these integrals?
1) x^2 dx / (x^2+x2)
2) dx / (x^3+9x)
3) dx / (x^4a^4)
4) (x^2+1) / (x^3+8)
And can you explain why please? Thank you :)
anonymous
 5 years ago
Integration: Out of the following methods, substitution/ parts/ partial fractions, which methods would you choose to solve each of these integrals? 1) x^2 dx / (x^2+x2) 2) dx / (x^3+9x) 3) dx / (x^4a^4) 4) (x^2+1) / (x^3+8) And can you explain why please? Thank you :)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, these are too hard for me :( I wish lokisan or sstarica would be logged in, I am sure they would be able to help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, thanks anyway. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For 1, note \[\frac{x^2}{x^2+x2} = \frac{(x^2+x2)(x2)}{x^2+x2} = 1\frac{x2}{(x+2)(x1)}\] then split it up into partial fraction (IMO) 2 is a partial fractions jobby. The other two would probably work with partial fractions but (if you're lucky) I'll try and see what is best.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow, this nooby site cut of the last denominator, but hopefully you can see what it is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you explain what you did with the first one? Did you divide or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's called 'adding nothing' (or similar); it's essentially a slick version of long division, yes  if the numerator and denominator are of the same order ALWAYS divide. If I don't see anything else, partial fractions should work for the last two: 3 is difference of two squares (twice!), and for the last one it should work too, as long as you see a factor (think what (simple) number will make the denominator = 0))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, thank you! That helped me a lot! :)
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