Question

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

Answer 1

try \(u=\sqrt{x}\)

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

\[e^x = A (e^x + 1) + B (2e^x -1)\]

Answer 14

\[\Large e^x = A (e^x + 1) + B (2e^x -1)\]
\[\Large e^x = A*e^x + A + 2Be^x - B\]

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

Answer 15

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

Answer 16

do you see how to get A and B?

Answer 17

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

Answer 18

wait
i think i got it

Answer 19

you saw how I got

1 = A+2B
0 = A-B

right? or no?

Answer 20

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

Answer 21

you're very close, but not quite there

Answer 22

You should find that A = B = 1/3

Answer 23

also, it's 2e^x - 1

not 2e^x + 1

Answer 24

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

Answer 25

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