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

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4Hint: 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
 one year ago
Best ResponseYou've already chosen the best response.0thank you. i used \[u=2+\sqrt{2}\] and i got \[\frac{ 3 }{ 4 }(2+\sqrt{x)} ^3 + c\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large \frac{2}{3}(2+\sqrt{x})^{3}+C\] is correct

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Is your function: \(\dfrac{dx}{\sqrt{x}}\cdot (2+\sqrt{x})^4\) ?

zaxoanl
 one year ago
Best ResponseYou've already chosen the best response.0can you give me a hint on this question as well \[dx/(1+2e^xe^x)\]

zaxoanl
 one year ago
Best ResponseYou've already chosen the best response.0its \[dx/(\sqrt{x}(2+\sqrt{x})^4)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4first I would multiply top and bottom by \[\Large e^x\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4\[\large \frac{dx}{1+2e^xe^{x}}\] \[\large \frac{e^x*dx}{e^x*(1+2e^xe^{x})}\] \[\large \frac{e^x*dx}{e^x*1+e^x*2e^xe^x*e^{x}}\] \[\large \frac{e^x*dx}{e^x+2e^{2x}1}\] \[\large \frac{e^x*dx}{e^x+2(e^{x})^21}\] \[\large \frac{e^x*dx}{2(e^{x})^2+e^x1}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4Then 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)(2z1) and then use partial fraction decomposition

zaxoanl
 one year ago
Best ResponseYou've already chosen the best response.0yes i did that and i don't know what number to use to cancel out the e^x

zaxoanl
 one year ago
Best ResponseYou've already chosen the best response.0\[e^x = A (e^x + 1) + B (2e^x 1)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large e^x = A (e^x + 1) + B (2e^x 1)\] \[\Large e^x = A*e^x + A + 2Be^x  B\] 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 = AB

zaxoanl
 one year ago
Best ResponseYou've already chosen the best response.0sorry don't get how to solve for the x values

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4do you see how to get A and B?

zaxoanl
 one year ago
Best ResponseYou've already chosen the best response.0how would you know what numbers to substitute in to solve for a and b

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4you saw how I got 1 = A+2B 0 = AB right? or no?

zaxoanl
 one year ago
Best ResponseYou've already chosen the best response.0yes 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
 one year ago
Best ResponseYou've already chosen the best response.4you're very close, but not quite there

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4You should find that A = B = 1/3

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4also, it's 2e^x  1 not 2e^x + 1

zaxoanl
 one year ago
Best ResponseYou've already chosen the best response.0oh yea i wrote it right just that i typed it wrong 2e^x  1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.4\[\large \frac{e^x}{(2e^{x}1)(e^{x}+1)}=\frac{A}{2e^{x}1}+\frac{B}{e^{x}+1}\] \[\large \frac{e^x}{(2e^{x}1)(e^{x}+1)}=\frac{1/3}{2e^{x}1}+\frac{1/3}{e^{x}+1}\] \[\large \frac{e^x}{(2e^{x}1)(e^{x}+1)}=\frac{1}{3}\left(\frac{1}{2e^{x}1}+\frac{1}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}1)(e^{x}+1)}=\frac{1}{3}\left(\frac{12e^x+2e^x}{2e^{x}1}+\frac{1e^x+e^x}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}1)(e^{x}+1)}=\frac{1}{3}\left(\frac{(2e^x1)+2e^x}{2e^{x}1}+\frac{e^x+1e^x}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}1)(e^{x}+1)}=\frac{1}{3}\left(\frac{2e^x1}{2e^{x}1}+\frac{2e^x}{2e^{x}1}+\frac{e^x+1}{e^{x}+1}\frac{e^x}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}1)(e^{x}+1)}=\frac{1}{3}\left(1+\frac{2e^x}{2e^{x}1}+1\frac{e^x}{e^{x}+1}\right)\] \[\large \frac{e^x}{(2e^{x}1)(e^{x}+1)}=\frac{1}{3}\left(\frac{2e^x}{2e^{x}1}\frac{e^x}{e^{x}+1}\right)\] I'll let you finish up
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