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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    Let's spice up this thread (plus I need an excuse to not do S3):

    Integrate \displaystyle \int e^{x+e^{x}} dx , hence or otherwise integrate \displaystyle \int e^{2x+e^{x}} dx

    If this is obvious to you:
    Spoiler:
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    Integrate again by considering the derivative of e^{x+e^{x}}
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    For the first part, is this it?

    (2/(x^2 +e^x))e^(x+e^x) +c
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    (Original post by lordoftheties)
    For the first part, is this it?

    (2/(x^2 +e^x))e^(x+e^x) +c
    Can you post your method? You've missed something, how did you solve it?
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    @Zacken can I learn M1 in 10 days?

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    (Original post by KINGYusuf)
    @Zacken can I learn M1 in 10 days?

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    You can learn M1 in less than a day with ExamSolutions
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    (Original post by KINGYusuf)
    @Zacken can I learn M1 in 10 days?

    Posted from TSR Mobile
    As above, it's very easy to learn the content, but you might take some time practicing your intuition and technique (when to resolve, how to resolve, etc...) but 10 days is more than enough.
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    Newtons 3rd law will save your ass for M1


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    (Original post by Euclidean)
    Let's spice up this thread (plus I need an excuse to not do S3):

    Integrate \displaystyle \int e^{x+e^{x}} dx , hence or otherwise integrate \displaystyle \int e^{2x+e^{x}} dx

    If this is obvious to you:
    Spoiler:
    Show
    Integrate again by considering the derivative of e^{x+e^{x}}
    For the first bit:
    Spoiler:
    Show
     e^{x+e^{x}} = e^x\cdot e^{e^{x}}\\ e^x =  e^{ln e^{x}}\\e^{x+e^{x}} = e^{ln e^{x}}\cdot  e^{e^{x}}\\ let\ u =e^x\\\frac{dx}{du} = \frac{1}{e^x}\\ \frac{dx}{du} = \frac{1}{u}\\\\ \int e^{x+e^{x}} dx = \int e^{ln u}\cdot e^u\cdot \frac{1}{u}\ du \\ \int e^{x+e^{x}} dx =\int\frac{u}{u}\cdot e^u\ du\\\int e^u\du = e^u + C\\ = e^{e^{x}} +C
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    (Original post by Euclidean)
    Let's spice up this thread (plus I need an excuse to not do S3):

    Integrate \displaystyle \int e^{x+e^{x}} dx , hence or otherwise integrate \displaystyle \int e^{2x+e^{x}} dx

    If this is obvious to you:
    Spoiler:
    Show
    Integrate again by considering the derivative of e^{x+e^{x}}
    Agree with NotNotBatman with the first one.

    Second one is e^{x+e^{x}}-e^{e^{x}} + C

    Same method but this time you have \displaystyle \int ue^{u}\ du\ \ \
    where u=e^{x}

    Which you can solve by parts
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    (Original post by NotNotBatman)
    For the first bit:
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     e^{x+e^{x}} = e^x\cdot e^{e^{x}}
    You could just note that that's \int f'(x) e^{f(x)} \, \mathrm{d}x = e^{f(x)}.
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    (Original post by Zacken)
    You could just note that that's \int f'(x) e^{f(x)} \, \mathrm{d}x = e^{f(x)}.
    oh yeah.
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    (Original post by edothero)
    Agree with NotNotBatman with the first one.

    Second one is e^{x+e^{x}}-e^{e^{x}} + C

    Same method but this time you have \displaystyle \int ue^{u}\ du\ \ \
    where u=e^{x}

    Which you can solve by parts
    Alternatively:
    Spoiler:
    Show

    \displaystyle \frac{d(e^{x+e^{x}})}{dx} = e^{x+e^{x}} + e^{x}e^{x+e^{x}} = e^{x+e^{x}} + e^{2x+e^{x}}

    Integrating both sides with respect to x:

    \displaystyle e^{x+e^{x}} + C = \int e^{x+e^{x}} dx + \int e^{2x+e^{x}} dx

    \displaystyle e^{x+e^{x}} + C = e^{e^{x}} + \int e^{2x+e^{x}} dx

    \displaystyle \int e^{2x+e^{x}} dx = e^{x+e^{x}} - e^{e^{x}} + C
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    (Original post by Euclidean)
    Alternatively:
    Spoiler:
    Show

    \displaystyle \frac{d(e^{x+e^{x}})}{dx} = e^{x+e^{x}} + e^{x}e^{x+e^{x}} = e^{x+e^{x}} + e^{2x+e^{x}}

    Integrating both sides with respect to x:

    \displaystyle e^{x+e^{x}} + C = \int e^{x+e^{x}} dx + \int e^{2x+e^{x}} dx

    \displaystyle e^{x+e^{x}} + C = e^{e^{x}} + \int e^{2x+e^{x}} dx

    \displaystyle \int e^{2x+e^{x}} dx = e^{x+e^{x}} - e^{e^{x}} + C
    That's quite a nice way of solving it actually..
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    Name:  _IYGB 5.jpg
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    http://www.thestudentroom.co.uk/show....php?t=3361905
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    Glad to have you back
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    Concerned that I just decided to spend part of my Friday night doing that integral! Exams are sending me mad...
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    (Original post by edothero)
    Glad to have you back
    I am not back.
    I cannot stand this site.
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    (Original post by edothero)
    Glad to have you back
    I second this!
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    Mercy please
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    (Original post by TeeEm)
    I am not back.
    I cannot stand this site.
    Well, the help you put forward to students during the exam season is definitely appreciated, thank you nevertheless
 
 
 
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