## zaxoanl one year ago need help on integrate dx / (sqrt x)(2+ sqrtx)^4

1. anonymous

try $$u=\sqrt{x}$$

2. jim_thompson5910

Hint: Let $$\LARGE u = 2+\sqrt{x}$$ which means $$\LARGE \frac{du}{dx} = \frac{1}{2\sqrt{x}}$$ which becomes $$\LARGE 2du = \frac{dx}{\sqrt{x}}$$

3. zaxoanl

thank you. i used $u=2+\sqrt{2}$ and i got $\frac{ -3 }{ 4 }(2+\sqrt{x)} ^-3 + c$

4. zaxoanl

i mean -2/3 (2+.,,,,

5. jim_thompson5910

$\Large -\frac{2}{3}(2+\sqrt{x})^{-3}+C$ is correct

6. Jhannybean

Is your function: $$\dfrac{dx}{\sqrt{x}}\cdot (2+\sqrt{x})^4$$ ?

7. zaxoanl

can you give me a hint on this question as well $dx/(1+2e^x-e^-x)$

8. zaxoanl

its $dx/(\sqrt{x}(2+\sqrt{x})^4)$

9. jim_thompson5910

first I would multiply top and bottom by $\Large e^x$

10. jim_thompson5910

$\large \frac{dx}{1+2e^x-e^{-x}}$ $\large \frac{e^x*dx}{e^x*(1+2e^x-e^{-x})}$ $\large \frac{e^x*dx}{e^x*1+e^x*2e^x-e^x*e^{-x}}$ $\large \frac{e^x*dx}{e^x+2e^{2x}-1}$ $\large \frac{e^x*dx}{e^x+2(e^{x})^2-1}$ $\large \frac{e^x*dx}{2(e^{x})^2+e^x-1}$

11. jim_thompson5910

Then notice how the denominator is of the form 2z^2 + z - 1 where z = e^x factor 2z^2 + z - 1 to get (z+1)(2z-1) and then use partial fraction decomposition

12. zaxoanl

yes i did that and i don't know what number to use to cancel out the e^x

13. zaxoanl

$e^x = A (e^x + 1) + B (2e^x -1)$

14. jim_thompson5910

$\Large e^x = A (e^x + 1) + B (2e^x -1)$ $\Large e^x = A*e^x + A + 2Be^x - B$ Grouping up the terms, we see that 1*e^x = A*e^x+2Be^x 1*e^x = e^x(A+2B) 1 = A+2B and 0 = A-B

15. zaxoanl

sorry don't get how to solve for the x values

16. jim_thompson5910

do you see how to get A and B?

17. zaxoanl

how would you know what numbers to substitute in to solve for a and b

18. zaxoanl

wait i think i got it

19. jim_thompson5910

you saw how I got 1 = A+2B 0 = A-B right? or no?

20. zaxoanl

yes i looked at something wrong there you were right so i got $\ln \left| 2e^x+1 \right|+\ln \left| e^x+1 \right|+c$

21. jim_thompson5910

you're very close, but not quite there

22. jim_thompson5910

You should find that A = B = 1/3

23. jim_thompson5910

also, it's 2e^x - 1 not 2e^x + 1

24. zaxoanl

oh yea i wrote it right just that i typed it wrong 2e^x - 1

25. jim_thompson5910

$\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{A}{2e^{x}-1}+\frac{B}{e^{x}+1}$ $\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1/3}{2e^{x}-1}+\frac{1/3}{e^{x}+1}$ $\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(\frac{1}{2e^{x}-1}+\frac{1}{e^{x}+1}\right)$ $\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(\frac{1-2e^x+2e^x}{2e^{x}-1}+\frac{1-e^x+e^x}{e^{x}+1}\right)$ $\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(\frac{-(2e^x-1)+2e^x}{2e^{x}-1}+\frac{e^x+1-e^x}{e^{x}+1}\right)$ $\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(-\frac{2e^x-1}{2e^{x}-1}+\frac{2e^x}{2e^{x}-1}+\frac{e^x+1}{e^{x}+1}-\frac{e^x}{e^{x}+1}\right)$ $\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(-1+\frac{2e^x}{2e^{x}-1}+1-\frac{e^x}{e^{x}+1}\right)$ $\large \frac{e^x}{(2e^{x}-1)(e^{x}+1)}=\frac{1}{3}\left(\frac{2e^x}{2e^{x}-1}-\frac{e^x}{e^{x}+1}\right)$ I'll let you finish up