roselin 2 years ago evaluate indefinite integrals integral squareroot x^4/ x^3-1 dx

1. roselin

what is it ?

2. roselin

thanks

is the x^3-1 in the sqrt also or no ??

4. roselin

yes, it is.

|dw:1355676028137:dw|

6. roselin

okay,

now, use int by subs let u=x^3-1 du = 3x^2 dx or x^2 dx = 1/3 du continue it.....

8. roselin

-1/3 intergral 1/ sqrt of u * du

miss (-) sign, how come this ?

10. roselin

i took du as -3x^2

derivative of x^3 is 3x^2, right ?

12. roselin

yes

so, why becomes (-) ?

14. roselin

well, i wrote the problem as 1-x^3 instead of x^3 -1

lol, hahaha...

16. roselin

hmm

so, the original problem is over x^3-1 or 1-x^3 ?

18. roselin

it is x^3 -1

ok, therefore the integration can be |dw:1355676826761:dw|

i just gave u +c :), for the rest let try it

21. roselin

u= x^3 -1 here

yes

because i have saw ur problem is .... over x^3-1, but experiment gave u 1-x^3

24. roselin

i know

25. roselin

and i put it in the wolfram thing, got some different ans

what's the result of int above, dont use wolfram first

27. roselin

(u) ^-1/2 +1/ -1/2+1

where is the 1/3 ?

29. roselin

(x^3-1)+c

30. roselin

What do I do with that 1/3?

31. roselin

can you please write down the whole problem and show how it works, if you do not mind then

32. roselin

i am kind of confused

|dw:1355677696473:dw|

got it ?

35. roselin

how did u get that 2u?

1/(-1/2 + 1) = 1/(1/2) = 1 * 2/1 = 2 right ?

37. roselin

yeah, sorry

38. roselin

i have another problem, will you guide me through that one? i will try solving it and show it to you.

nopes... now, subtitute back that u=x^3-1 so, the answer be 2/3 (x^3-1)^(1/2) + c or 2/3*sqrt(x^3-1) + c

wolfram is same with the answer above :)

41. roselin

k.

42. roselin

okay

43. roselin

thank you.