Question

Can anyone help me with subtle U-substitution??
integral of e^(2z)/e^(2z)-4e^(-2z) dz

Answer 1

\[\int\limits \frac{ e ^{2z} }{ e^{2z}-4e^{-2z} } dz\]

Answer 2

take e^2z as x

Answer 3

as x?

Answer 4

yeah watever you are comfortable with

Answer 5

ya i like that idea.
Confused on how it will work out rod?

Answer 6

would the substitution
\(\Large e^{4z} = x\) after multiplying numerator and denominator by \(e^{2x}\)
make the solution easier?

Answer 7

Yes, I'm not sure where I would go from there

Answer 8

i meant ....by \(e^{2z}\)

Answer 9

\[\large\rm x=e^{2z}\qquad\qquad\qquad\frac{1}{x}=e^{-2z}\]Yah multiplying through is another option :)

Answer 10

why not? exactly zepdrix. youll get integral 1/(x^2+2^2) form

Answer 11

with the substitution \(x= e^{2z}\), you will get a quadratic term in the denominator,
with the substitution \(x= e^{4z}\), you will get a linear term in the denominator,

Answer 12

but then what about the numerator? it will be like integral x^2/(x+4)?

Answer 13

You figure this one out ruby rod? :)

Answer 14

@random231 the numerator will be \(e^{4z} dz\)
which will actually be dx

so your new integral will be of the form
\(\dfrac{dx}{x-4}\)

Answer 15

mhmm yeah hartnn that wub be a lot easier!

Answer 16

I think I'm starting to understand it, I'm only wondering what it was multiplied by to simplify the denominator that way?

Answer 17

denominator:
\(\Large e^{2z}-4e^{-2z} = e^{2z}-4{\dfrac{1}{e^{2z}}} = \dfrac{e^{4z}-4}{e^{2z}}\)

Answer 18

ohhh ok, now I see it