Question

whats the integral for x^4 / (1-x) ??

Answer 1

\[\huge \int \frac{x^4}{1-x}dx\]

Answer 2

yep, but how can I solve it?

Answer 3

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...

Answer 4

what u mean with rational expression?

Answer 5

Fraction. Fancy word for fraction.

Answer 6

so u say that Ive to use u=x^4, du/dx=4x^3 so dx=du/4x^3
then: \[\frac{ u }{ 1-x }\frac{ du }{ 4x^3 }\]

Answer 7

??

Answer 8

Unfortunately not that simple. u was in your denominator, wasn't it?

Answer 9

oh yeep ure right, so its:
u=(1-x)
du/dx= -1
dx= du/-1
and then:
\[\int\limits_{}^{}\frac{ x^4 }{ u}*\frac{ du }{ -1 } = -\int\limits_{}^{}\frac{ x^4 }{ u}*du\]

Answer 10

??

Answer 11

Okay, much better. But you cannot solve this integral without expressing \(x^4\) in terms of u.

Answer 12

so.. what can I do??

Answer 13

Well
\[\large u = 1-x\]
\[\large x = 1-u\]

\[\huge x^4 = (1-u)^4\]

Answer 14

wow! so uhm,, what do I have to do now?

Answer 15

Expand.

Answer 16

\[\large (1-u)^4 = u^4 -4u^3 +6u^2 -4u +1\]

Answer 17

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

Answer 18

No...
remember, you started with
\[\huge \int \frac{x^4}{1-x}dx\]And you let u = 1-x, work from there, and substitute.

Answer 19

mm so what I = to "u" then? :/

Answer 20

Shun being spoonfed, @appleduardo ... :P
\[\large u = 1-x\]\[\large du = -dx\]\[\large dx = -du\]\[\large x = 1-u\]\[\large x^4=(1-u)^4\]