Find the antiderivative

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Homework Statement


Find the antiderivative of each derivative:

dp/dx = [(e^x)+(e^-x)]/[((e^x)-(e^-x))^2]


The Attempt at a Solution



I can't entirely figure out how to get myself started here.
I tired splitting up the derivative so that it's

[(e^x)/ (((e^x)-(e^-x))^2)] + [(e^-x)/(((e^x)-(e^-x))^2)]

[(e^x)((e^x)-(e^-x))^-2] + [(e^-x)((e^x)-(e^-x))^-2]

I think I should try substitution but I can't figure out what value to make u.

If anyone could help me, that would be fantastic. I don't think I'm going quite in the right direction.
Thanks!
 

Answers and Replies

  • #2
Dick
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This would be a good time to learn about hyperbolic function like sinh(x) and cosh(x), but you can also do it directly. Take u=e^x. So e^(-x)=1/u.
 
  • #3
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I would suggest thinking about sinh and cosh... and then relying on some trigonometric formulas. I am not good with these hyperbolic things.

edit: oops.. dick already mentioned it while I was solving using those sinhs...
 
  • #4
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Thanks. Technically, for this exam I'm studying for, we're not supposed to know those hyperbolic functions- so I've got to make use of substitution.

Okay, so:

I split the derivative into the two parts:(e^x)[(e^x)-(e^-x)]^-2 and (e^-x)[(e^x)-(e^-x)]^-2 and made u=e^x so du=(e^x)dx and e^-x=1/e^x

So the antiderivative of (e^x)[(e^x)-(e^-x)]^-2 = [-1/u]-[(u^3)/3]

so now I'm looking for the antiderivative of (e^-x)[(e^x)-(e^-x)]^-2

So I end up with [(4(u^3))/3]-4ln|u|

Which makes the antiderivative=
[-1/u]-[(u^3)/3]+[(4(u^3))/3]-4ln|u|
then I change the u's back.

[-1/(e^x)]-[((e^x)^3)/3]+[(4((e^x)^3))/3]-4ln|(e^x)|

so I get [-1/(e^x)]-[(e^3x)/3]+[(4(e^3x))/3]-4x

it gives me: [-e^-x]-[e^3x]-x

...and the book tells me I'm still wrong.

Can anyone help me out??
thanks.
 
  • #5
Dick
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Thanks. Technically, for this exam I'm studying for, we're not supposed to know those hyperbolic functions- so I've got to make use of substitution.

Okay, so:

I split the derivative into the two parts:(e^x)[(e^x)-(e^-x)]^-2 and (e^-x)[(e^x)-(e^-x)]^-2 and made u=e^x so du=(e^x)dx and e^-x=1/e^x

So the antiderivative of (e^x)[(e^x)-(e^-x)]^-2 = [-1/u]-[(u^3)/3]

You are already way off track. I don't see how you got that antiderivative. Just do the substitution without splitting anything up. Show the whole integral in terms of u before you go leaping forward.
 
  • #6
Dick
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oh. okay

[tex]\int(e^{x}+e^{-x})(e^{x}-e^{-x})^{-2}dx[/tex]

let u=e[tex]^{x}[/tex]
du=e[tex]^{x}[/tex] ; du=dx

e[tex]^{-x}[/tex] =1/e[tex]^{x}[/tex]

okay so from there:

[tex]\int(u+1/u)(u-1/u)^{-2}du[/tex]

[tex]\int(u+1/u)(u^{-2}-u^{2})du[/tex]

That far at least is okay, I hope.

Stop there. (u-1/u)^(-2) is not equal to (u^(-2)-u^(2)). You can't distribute powers like that. You'll never get to the end correctly with algebra like that. And dx is not equal to du. du=e^(x)*dx=u*dx.
 
  • #7
Dick
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I told you. You are doing bad algebra. (u-1/u)^(-2) is not equal to (u^(-2)-u^(2)). If you don't know what to do with it, don't do anything with it.
 
Last edited:
  • #8
Dick
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I'll give you a hint. (u+1/u)=(u^2+1)/u. (u-1/u)^2=((u^2-1)/u)^2=(u^2-1)^2/u^2. I'm trying to write these expressions as polynomials in u so I can factor them and use partial fractions.
 
  • #9
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I'll give you a hint. (u+1/u)=(u^2+1)/u. (u-1/u)^2=((u^2-1)/u)^2=(u^2-1)^2/u^2. I'm trying to write these expressions as polynomials in u so I can factor them and use partial fractions.

Thanks Dick, you're really helping me right now. I have a lot of trouble with math and so I'm getting very turned around on this problem.

That said, I don't really understand your hint. I'm just getting really frustrated and confused with this.

I'm going to try to go through it step for step. Maybe you (or anyone) can point out to me
where that hint could help me?

so:
I'm looking for the antiderivative of [(e^x)+(e^-x)]/[(e^x)-(e^-x)]^2

and so u=e^x
du=(e^x)dx ; du=udx
e^-x = 1/e^x

okay... I= [[(u)+(1/u)]/[(u)-(1/u)]^2]dx
because there's a u in the numerator can I combine that with the dx to make the du?

I= [(u^-1)/[u-(u^-1)]^2] du

Okay- so I'm going to look at the denominator for a second:
I have to factor that out, right? to (u^2)-2-(1/u^2)

I= (1/u)/[(u^2)-2-(1/u^2)] du

can I multiply [(1/u^2)-(1/2)-u^2)] by (1/u) ?
is this closer to the right track???

I'm really sorry I sound so inept, I just don't understand and I am so confused!

Any help would be a life-saver. Thanks.
 
  • #10
HallsofIvy
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Thanks Dick, you're really helping me right now. I have a lot of trouble with math and so I'm getting very turned around on this problem.

That said, I don't really understand your hint. I'm just getting really frustrated and confused with this.

I'm going to try to go through it step for step. Maybe you (or anyone) can point out to me
where that hint could help me?

so:
I'm looking for the antiderivative of [(e^x)+(e^-x)]/[(e^x)-(e^-x)]^2

and so u=e^x
du=(e^x)dx ; du=udx
e^-x = 1/e^x

okay... I= [[(u)+(1/u)]/[(u)-(1/u)]^2]dx
because there's a u in the numerator can I combine that with the dx to make the du?

I= [(u^-1)/[u-(u^-1)]^2] du
No, you certainly can't do that- the "u" you are using is added, not multiplied What you could do that might help is multiply both numerator and denominator by u2.

Okay- so I'm going to look at the denominator for a second:
I have to factor that out, right? to (u^2)-2-(1/u^2)

I= (1/u)/[(u^2)-2-(1/u^2)] du

can I multiply [(1/u^2)-(1/2)-u^2)] by (1/u) ?
is this closer to the right track???

I'm really sorry I sound so inept, I just don't understand and I am so confused!

Any help would be a life-saver. Thanks.
 
  • #11
Dick
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What's making you sound inept is obvious algebraic mistakes. Don't make them. I'll give you a target to shoot for. The integrand becomes (u^2+1)*du/(u^2-1)^2 if you play your cards right. That was supposed to be the easy part. Now you have to use partial fractions on that. Which also needs some algebra skills without making mistakes on every line. Try the first one first, ok?
 
  • #12
dextercioby
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Try the substitution

[tex] u=e^{x}-e^{-x} [/tex].

What is "du" equal to ??
 
  • #13
Dick
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Try the substitution

[tex] u=e^{x}-e^{-x} [/tex].

What is "du" equal to ??

Good idea. Much less work.
 
  • #14
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Ah thank you. I finally managed to figure it out.

dp/dx = [(e^x)+(e^-x)]/[((e^x)-(e^-x))^2]

u=e^x - e^-x
du= e^x - e^-x (-1) dx
du= - e^x + e^-x dx

I= 1/u^2 (du)
I= u^-2 (du)
I= -1 u^-1
I= 1/-u + c

p= -1/( e^x - e^-x) + C


Thanks for all the suggestions and help- I definitely just hit a ridiculous mental block with this paticular question. Thanks again!
 

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