## anonymous 4 years ago integrate sqrt((x^3 -3)/x^11)

1. anonymous

|dw:1327888347751:dw|

2. anonymous

?

3. anonymous

maybe we can take out that annoying $x^{11}$ in the denominator and write is as $\frac{1}{x^5}\sqrt{\frac{x^3-3}{x}}$

4. anonymous

no that doesn't work. hmmm

5. anonymous

the answer in the back of the book is 2/27 (1-3/x^3)^(3/2)+C

6. anonymous

i got the answer from wolfram, but not the "show steps" and the answer makes it look like a u - sub so maybe we can arrange it to be one

7. anonymous

it is supposed to be solved by u substitution

8. anonymous

ok i have another idea. write it as $\frac{1}{x^4}\sqrt{\frac{x^3-3}{x^3}}$ and then make $u=\frac{x^3-3}{x^3}$

9. anonymous

that will do it.

10. anonymous

okay, thankyou so much

11. anonymous

let me try that

12. anonymous

yw let me know if you run in to any problems, but a miracle will occur when you find $du$ it will be just what you want

13. anonymous

okay lol for sure.

14. anonymous

it gives me (9/x^7) as du.. :/

15. anonymous

check again, should be $\frac{9}{x^4}$ as needed

16. anonymous

ok i am gunna one more time.

17. anonymous

$\frac{d}{dx}\frac{x^3-3}{x^3}=\frac{x^3\times 3x^2-3x^2\times (x^3-3)}{x^6}$

18. anonymous

all the $x^5$ terms go, you get $\frac{9x^2}{x^6}=\frac{9}{x^4}$

19. anonymous

in fact you cannot get a power of 7 in the denominator, because $(x^3)^2=x^6$

20. anonymous

oh shoot, i see my mistake, thankyou.. i was trying to do it with product rule

21. anonymous

everyone thinks that is the easy way, but believe me it is not. even wolfram does it that way, but then you have to add the fractions an no one is good at that

22. anonymous

yeah you are right, i will try not to use it, btw thankyou so much, ur a life saver

23. myininaya

$\frac{d}{dx}(\frac{x^3-3}{x^3})=\frac{d}{dx}(1-\frac{3}{x^3})$ quotient rule not needed but good work (the credit for this little part goes to JamesJ) Great work on figuring out the substitution, satellite. You're awesome! :)

24. anonymous

$\color{red}{\text{blush}}$