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appleduardo
whats the integral for x^4 / (1-x) ??
\[\huge \int \frac{x^4}{1-x}dx\]
yep, but how can I solve it?
It's actually quite easy, but incredibly tedious. Use u-substitution. When in doubt, attempt to let u = the denominator of a rational expression... chances are, that's the one...
what u mean with rational expression?
Fraction. Fancy word for fraction.
so u say that Ive to use u=x^4, du/dx=4x^3 so dx=du/4x^3 then: \[\frac{ u }{ 1-x }\frac{ du }{ 4x^3 }\]
Unfortunately not that simple. u was in your denominator, wasn't it?
oh yeep ure right, so its: u=(1-x) du/dx= -1 dx= du/-1 and then: \[\int\limits_{}^{}\frac{ x^4 }{ u}*\frac{ du }{ -1 } = -\int\limits_{}^{}\frac{ x^4 }{ u}*du\]
Okay, much better. But you cannot solve this integral without expressing \(x^4\) in terms of u.
so.. what can I do??
Well \[\large u = 1-x\] \[\large x = 1-u\] \[\huge x^4 = (1-u)^4\]
wow! so uhm,, what do I have to do now?
\[\large (1-u)^4 = u^4 -4u^3 +6u^2 -4u +1\]
so now I have: \[\frac{ x^4 }{ u^4 - 4u^3 + 6u^2 - 4u + 1 }\] but at this point is u still = to (1-x) ? sorry if this sounds silly or so, but I got a little confused when u got (1-u)^4
No... remember, you started with \[\huge \int \frac{x^4}{1-x}dx\]And you let u = 1-x, work from there, and substitute.
mm so what I = to "u" then? :/
Shun being spoonfed, @appleduardo ... :P \[\large u = 1-x\]\[\large du = -dx\]\[\large dx = -du\]\[\large x = 1-u\]\[\large x^4=(1-u)^4\]