C3 intergration by substitution, help me please! Watch

nevagonafail
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How do i intergrate\int \frac{2}{e^2^x+4}} dx given that u = e^2^x+4
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nevagonafail
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Someone please help me, im going to bed soon. I thought TSR was full of geniuses.
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nevagonafail
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So no ones got the balls to answer this question. Im going to bed now hopefully you guys or i will figure it out by tomorrow.
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giran
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int 2/u dx

u = e^2x + 4, 0.5 ln (u-4) = x
du/dx = 2e^2x

therefore dx = du/2e^2x

Find the integral of 2 * 2e^2(0.5 ln (u-4)) / u
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hai2410
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(Original post by nevagonafail)
So no ones got the balls to answer this question.
'No ones got the balls' haha. It was just that nobody could be arsed
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nuodai
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(Original post by nevagonafail)
So no ones got the balls to answer this question. Im going to bed now hopefully you guys or i will figure it out by tomorrow.
I'm sorry, remind me to reply instantly next time so as not to anger your superior manhood. We "geniuses of TSR" are of course a disposable resource that don't require any 'pleases' or 'thank yous', and deserve being insulted when we're not instant to suffice your demands; I mean, what are manners on the internet anyway?
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greenjm90
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(Original post by nuodai)
I'm sorry, remind me to reply instantly next time so as not to anger your superior manhood. I am of course a disposable resource that doesn't require any 'pleases' or 'thank yous', and deserve being insulted when I'm not instant to suffice your demands; I mean, what are manners on the internet?

Rude...
This.

Although I will say that what you've posted doesn't seem to be a very nice integral and I would have thought would be beyond the realms of C3. Where did you get it from?
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Unbounded
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(Original post by nevagonafail)
So no ones got the balls to answer this question. Im going to bed now hopefully you guys or i will figure it out by tomorrow.
kthxbai
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 u = e^{2x}+4 \implies \dfrac{\mathrm{d}u}{\mathrm{d}x} = 2e^{2x} = 2(u - 4) \iff \mathrm{d}x = \dfrac{1}{2(u-4)}\mathrm{d}u

 I = \displaystyle\int \dfrac{2}{e^{2x}+4}\ \mathrm{d}x = \displaystyle\int \dfrac{1}{u(u-4)}\ \mathrm{d}u

 I = \dfrac{1}{4}\displaystyle\int \left( \dfrac{1}{u-4} - \dfrac{1}{u}\right) \ \mathrm{d}u

 I = \frac{1}{4}\ln |u-4| - \frac{1}{4}\ln |u| + C = \frac{1}{2}x - \frac{1}{4}\ln (e^{2x}+4) + C
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nevagonafail
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:eek:
(Original post by greenjm90)
This.

Although I will say that what you've posted doesn't seem to be a very nice integral and I would have thought would be beyond the realms of C3. Where did you get it from?

First of all, apologies fore my behaviour it just that i was annoyed with that question. So please can someone help me out. thanks in advance.


Anyway in response to your question, its from the heinnemann AQA book.
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nevagonafail
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(Original post by GHOSH-5)
kthxbai
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 u = e^{2x}+4 \implies \dfrac{\mathrm{d}u}{\mathrm{d}x} = 2e^{2x} = 2(u - 4) \iff \mathrm{d}x = \dfrac{1}{2(u-4)}\mathrm{d}u

 I = \displaystyle\int \dfrac{2}{e^{2x}+4}\ \mathrm{d}x = \displaystyle\int \dfrac{1}{u(u-4)}\ \mathrm{d}u

 I = \dfrac{1}{4}\displaystyle\int \left( \dfrac{1}{u-4} - \dfrac{1}{u}\right) \ \mathrm{d}u

 I = \frac{1}{4}\ln |u-4| - \frac{1}{4}\ln |u| + C = \frac{1}{2}x - \frac{1}{4}\ln (e^{2x}+4) + C
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lol that was genuinely funny.
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Lou Reed
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(Original post by nevagonafail)
How do i intergrate\int \frac{2}{e^2^x+4}} dx given that u = e^2^x+4
(Original post by nevagonafail)
How do i intergrate\int \frac{2}{e^2^x+4}} dx given that u = e^2^x+4
youre given...

\int \frac{2}{e^{2x}+4}} \cdot dx

and...

u = e^{2x}+4

rearrange that to get..

 x = \frac{1}{2}\ln(u-4)

then differentiate it, so...

 \frac{du}{dx} = 2e^{2x}

separate the variables and substitute in x.

\int dx = \int \frac{1}{2e^{\ln(u-4)}} \cdot du

so....

 \int \frac{2}{u2e^{\ln(u-4)}} \cdot du

 = \int \frac{2}{2u(u-4)} \cdot du

 = \int \frac{1}{u(u-4)} \cdot du

 = \int \frac{1}{4u-16} - \frac{1}{4u}

 = \frac{1}{4}\ln|\frac{u-4}{u}| + C

subbing in u = f(x)...

 \frac{1}{4}\ln|\frac{e^{2x}}{e^{  2x}+4}| + C

hope thats right
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giran
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(Original post by Lou Reed)
youre given...

\int \frac{2}{e^{2x}+4}} \cdot dx

and...

u = e^{2x}+4

rearrange that to get..

 x = \frac{1}{2}\ln(u-4)

then differentiate it, so...

 \frac{du}{dx} = 2e^{2x}
[/latex]

hope thats right
Up to there I see what you've done is right, but I've never seen anyone seperate the variable for this?

You only need to seperate the variables when you have to integrate a function with both x and y's.
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Lou Reed
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(Original post by giran)
Up to there I see what you've done is right, but I've never seen anyone seperate the variable for this?

You only need to seperate the variables when you have to integrate a function with both x and y's.
no im pretty sure you need to separate the variables here to get (int)dx in terms of (int)du.

anyways i got the right answer =]
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giran
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(Original post by Lou Reed)
no im pretty sure you need to separate the variables here to get (int)dx in terms of (int)du.

anyways i got the right answer =]
That's stange, normally you'd make it in terms of U hence substitution. integrate with respect to U then stick u = whatever.

answer shouldn't be that.
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Lou Reed
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(Original post by giran)
That's stange, normally you'd make it in terms of U hence substitution. integrate with respect to U then stick u = whatever.

answer shouldn't be that.
i thnk were arguing the same point here..

we need to turn

\int \frac{2}{e^{2x}+4}} \cdot dx

into terms of u, ye?

so i need to convert the \int dx into terms of \int du so i can integrate with respect to u.

so...

u = e^{2x}+4

 \frac{du}{dx} = 2e^{2x}

\int dx = \int \frac{1}{2e^{\ln(u-4)}} \cdot du


btw, is my final integration wrong?
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nevagonafail
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(Original post by Lou Reed)
i thnk were arguing the same point here..

we need to turn

\int \frac{2}{e^{2x}+4}} \cdot dx

into terms of u, ye?

so i need to convert the \int dx into terms of \int du so i can integrate with respect to u.

so...

u = e^{2x}+4

 \frac{du}{dx} = 2e^{2x}

\int dx = \int \fra{1}{2e^{\ln(u-4)}} \cdot du


btw, is my final integration wrong?

Answer's meant to be
\frac{1}{2}x-\frac{1}{4}ln(e^2^x+4)+c
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giran
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(Original post by Lou Reed)
i thnk were arguing the same point here..

we need to turn

\int \frac{2}{e^{2x}+4}} \cdot dx

into terms of u, ye?

so i need to convert the \int dx into terms of \int du so i can integrate with respect to u.

so...

u = e^{2x}+4

 \frac{du}{dx} = 2e^{2x}

\int dx = \int \frac{1}{2e^{\ln(u-4)}} \cdot du


btw, is my final integration wrong?
you're confusing two different methods here.
The answer OP posted was the same one I got. Op that is correct.

 \frac{du}{2e^{2x}} = dx

Then you plug it back into your original question but in terms of u.
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Unbounded
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(Original post by nevagonafail)
lol that was genuinely funny.
Did you eventually find the solution, somewhere within the maze of the 59 spoilers? :awesome:
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nevagonafail
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(Original post by GHOSH-5)
Did you eventually find the solution, somewhere within the maze of the 59 spoilers? :awesome:
Yes, i did thats why i found it funny. Though I'd rather you give me the answer straight next time.
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Lou Reed
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(Original post by giran)
answer shouldn't be that.
(Original post by nevagonafail)
Answer's meant to be
\frac{1}{2}x-\frac{1}{4}ln(e^2^x+4)+c
yeah thats what i got...(from my previous post)... \frac{1}{4}\ln|\frac{e^{2x}}{e^{  2x}+4}| + C

 = \frac{1}{4}\ln e^{2x} - \frac{1}{4}\ln|e^{2x} + 4| + C

 = \frac{2x}{4} - \frac{1}{4}\ln|e^{2x} + 4| + C

 = \frac{1}{2}x - \frac{1}{4}\ln|e^{2x} + 4| + C

:awesome:
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