need help on integrate dx / (sqrt x)(2+ sqrtx)^4

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

need help on integrate dx / (sqrt x)(2+ sqrtx)^4

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

try \(u=\sqrt{x}\)
Hint: Let \(\LARGE u = 2+\sqrt{x}\) which means \(\LARGE \frac{du}{dx} = \frac{1}{2\sqrt{x}}\) which becomes \(\LARGE 2du = \frac{dx}{\sqrt{x}}\)
thank you. i used \[u=2+\sqrt{2}\] and i got \[\frac{ -3 }{ 4 }(2+\sqrt{x)} ^-3 + c\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question