Integration Problem (1 Viewer)

[_ShiFTy_]

I dont know the symbol inputs...

Whats the integration of (e^x)/(e^2x - 1 )

 

[_ShiFTy_]

hehe forgot that u^2 - 1 can be factorised

Will you also be able to use the formula

1/2a. (ln |(a+x)/(a-x)|)

When integrating 1/(a^2 - x^2)

 

[Riviet]

hehe forgot that u^2 - 1 can be factorised

Will you also be able to use the formula

1/2a. (ln |(a+x)/(a-x)|)

When integrating 1/(a^2 - x^2)

I personally wouldn't, as you taking a major shortcut in going straight to the answer and showing very little working or skill in doing the integration; I think you're best off doing it the normal way with A/(u+1) + B/(u-1) = 1/(u²-1)

 

[bboyelement]

our teacher taught us to do the shortcuts ... is that bad and also in the first exercise of cambridge (integration) it also shows you the shortcuts ... i still comtemplating whether to use the substitution or not ...

 

[Riviet]

our teacher taught us to do the shortcuts ... is that bad and also in the first exercise of cambridge (integration) it also shows you the shortcuts ... i still comtemplating whether to use the substitution or not ...

If you get it wrong, you won't get any marks, as you haven't given the marker any oppurtunity to award you some marks "for showing working". Also, if you come across a hard integral, your shortcuts usually won't work or will be too hard to apply to the integral.

 

[icycloud]

Letting e^x = sec gives an interesting (but perhaps longer) way of solving this integral without partial fractions .

I=∫e^x/(e^(2x)-1)dx
Let e^x = sec
e^x dx = sectan du
dx = tan du

I=∫sectan / tan^2 du
= ∫sec/tan du
= ∫cosecdu
= ∫(cosec^2+coseccot)/(cosec+cot)du
= -ln[cosec+cot] + c
= ln[sqrt(e^(2x)-1) / (e^x+1)] +c

(=ln[sqrt[(e^x-1)/(e^x+1)]] + c <-- which is the answer you get using partial fractions)

 

[_ShiFTy_]

Letting e^x = sec gives an interesting (but perhaps longer) way of solving this integral without partial fractions .

I=∫e^x/(e^(2x)-1)dx
Let e^x = sec
e^x dx = sectan du
dx = tan du

I=∫sectan / tan^2 du
= ∫sec/tan du
= ∫cosecdu
= ∫(cosec^2+coseccot)/(cosec+cot)du
= -ln[cosec+cot] + c
= ln[sqrt(e^(2x)-1) / (e^x+1)] +c

(=ln[sqrt[(e^x-1)/(e^x+1)]] + c <-- which is the answer you get using partial fractions)

Does that actually work??
Because i tried letting e^x = sinu, e^x dx = cosudu

I got up to:

I= -∫secu du
= -ln (secu + tanu) + c

Then i got stuck...

 

[icycloud]

Does that actually work??
Because i tried letting e^x = sinu, e^x dx = cosudu

I got up to:

I= -∫secu du
= -ln (secu + tanu) + c

Then i got stuck...

Of course.

- ln (secu+tanu) + c = -ln(1/cosu + sinu/cosu)+ c
= -ln( [sinu+1] / cosu)+ c
= -ln( [e^x+1] / sqrt(1-sin²u))+ c
= -ln([e^x+1]/sqrt[1-e^(2x)]) + c
= ln(sqrt[1-e^(2x)]/[e^x+1]) + c

 

[_ShiFTy_]

Of course.

- ln (secu+tanu) + c = -ln(1/cosu + sinu/cosu)+ c
= -ln( [sinu+1] / cosu)+ c
= -ln( [e^x+1] / sqrt(1-sin²u))+ c
= -ln([e^x+1]/sqrt[1-e^(2x)]) + c
= ln(sqrt[1-e^(2x)]/[e^x+1]) + c

thx man
I was doubting the validity of this method when i first started trying to do this question

 

[_ShiFTy_]

Did my 4U paper today and this was one of the questions

int. (e^x)/(9 + e^2x)

Very similar to the question i asked before...

Im not sure if im right, but i got

1/3 . tan-1 (e^x / 3)

Can someone please verify this

 

[Riviet]

I'm not sure how you got the 1/3's in the answer, but I got tan^-1e^x + C, by using a substition of u=e^x.

 

[_ShiFTy_]

I'm not sure how you got the 1/3's in the answer, but I got tan^-1e^x + C, by using a substition of u=e^x.

Oops, sorry the denominator was supposed to say 9 + e^2x

 

[Riviet]

Yeah, I got the same answer. Nice work.