## appleduardo 2 years ago whats the integral for x^4 / (1-x) ??

1. terenzreignz

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

2. appleduardo

yep, but how can I solve it?

3. terenzreignz

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...

4. appleduardo

what u mean with rational expression?

5. terenzreignz

Fraction. Fancy word for fraction.

6. appleduardo

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

7. appleduardo

??

8. terenzreignz

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

9. appleduardo

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

10. appleduardo

??

11. terenzreignz

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

12. appleduardo

so.. what can I do??

13. terenzreignz

Well $\large u = 1-x$ $\large x = 1-u$ $\huge x^4 = (1-u)^4$

14. appleduardo

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

15. terenzreignz

Expand.

16. terenzreignz

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

17. appleduardo

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

18. terenzreignz

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

19. appleduardo

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

20. terenzreignz

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$