zaxoanl
  • zaxoanl
need help on integrate dx / (sqrt x)(2+ sqrtx)^4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
try \(u=\sqrt{x}\)
jim_thompson5910
  • jim_thompson5910
Hint: Let \(\LARGE u = 2+\sqrt{x}\) which means \(\LARGE \frac{du}{dx} = \frac{1}{2\sqrt{x}}\) which becomes \(\LARGE 2du = \frac{dx}{\sqrt{x}}\)
zaxoanl
  • zaxoanl
thank you. i used \[u=2+\sqrt{2}\] and i got \[\frac{ -3 }{ 4 }(2+\sqrt{x)} ^-3 + c\]

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zaxoanl
  • zaxoanl
i mean -2/3 (2+.,,,,
jim_thompson5910
  • jim_thompson5910
\[\Large -\frac{2}{3}(2+\sqrt{x})^{-3}+C\] is correct
Jhannybean
  • Jhannybean
Is your function: \(\dfrac{dx}{\sqrt{x}}\cdot (2+\sqrt{x})^4\) ?
zaxoanl
  • zaxoanl
can you give me a hint on this question as well \[dx/(1+2e^x-e^-x)\]
zaxoanl
  • zaxoanl
its \[dx/(\sqrt{x}(2+\sqrt{x})^4)\]
jim_thompson5910
  • jim_thompson5910
first I would multiply top and bottom by \[\Large e^x\]
jim_thompson5910
  • jim_thompson5910
\[\large \frac{dx}{1+2e^x-e^{-x}}\] \[\large \frac{e^x*dx}{e^x*(1+2e^x-e^{-x})}\] \[\large \frac{e^x*dx}{e^x*1+e^x*2e^x-e^x*e^{-x}}\] \[\large \frac{e^x*dx}{e^x+2e^{2x}-1}\] \[\large \frac{e^x*dx}{e^x+2(e^{x})^2-1}\] \[\large \frac{e^x*dx}{2(e^{x})^2+e^x-1}\]
jim_thompson5910
  • 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
zaxoanl
  • zaxoanl
yes i did that and i don't know what number to use to cancel out the e^x
zaxoanl
  • zaxoanl
\[e^x = A (e^x + 1) + B (2e^x -1)\]
jim_thompson5910
  • 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
zaxoanl
  • zaxoanl
sorry don't get how to solve for the x values
jim_thompson5910
  • jim_thompson5910
do you see how to get A and B?
zaxoanl
  • zaxoanl
how would you know what numbers to substitute in to solve for a and b
zaxoanl
  • zaxoanl
wait i think i got it
jim_thompson5910
  • jim_thompson5910
you saw how I got 1 = A+2B 0 = A-B right? or no?
zaxoanl
  • zaxoanl
yes i looked at something wrong there you were right so i got \[\ln \left| 2e^x+1 \right|+\ln \left| e^x+1 \right|+c\]
jim_thompson5910
  • jim_thompson5910
you're very close, but not quite there
jim_thompson5910
  • jim_thompson5910
You should find that A = B = 1/3
jim_thompson5910
  • jim_thompson5910
also, it's 2e^x - 1 not 2e^x + 1
zaxoanl
  • zaxoanl
oh yea i wrote it right just that i typed it wrong 2e^x - 1
jim_thompson5910
  • 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

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