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Math Help - Integral , just check if i'm right plz

  1. #1
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    Integral , just check if i'm right plz

    \int (1+e^x)^{\frac{1}{2}} dx

    I did u^2 = 1+e^x

    I got 2\int  u^2/(u^2 -1)du

    Is this correct ?
    ------------------------------

    \int x^3/(x-1)^4 dx

    I used parts \int x^3(x-1)^-4 = (-1/3)(x^3/(x-1)^3 .....

    I know you could do u =x -1

    But is my way wrong ?
    ------------------------------------

    \int 1/(x)(4x^2 +1)^{\frac{1}{2}}dx
    i did

    u ^2 = (4x^2 +1)

    I got the integral to

    \int 1/(u^2-1)du

    Is this right ?
    ----------------------------------


    I just want you to tell me if i'm right because my book gives different answers when i do the integral my way.
    Last edited by Banned for attempted hacking; February 8th 2009 at 04:53 PM. Reason: wrote the question wrong. These questions are different there are 3 questions in total.
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  2. #2
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    Quote Originally Posted by ╔(σ_σ)╝ View Post
    \int (1+ex)^{\frac{1}{2}} dx
    why all the substitution?

    \int (1+ex)^{\frac{1}{2}} \, dx = \frac{2(1+ex)^{\frac{3}{2}}}{3e} + C

    did you mean ...

    \int (1+e^x)^{\frac{1}{2}} \, dx ?
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  3. #3
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    Quote Originally Posted by skeeter View Post
    why all the substitution?

    \int (1+ex)^{\frac{1}{2}} \, dx = \frac{2(1+ex)^{\frac{3}{2}}}{3e} + C

    did you mean ...

    \int (1+e^x)^{\frac{1}{2}} \, dx ?
    Yeah

    If it was e^x , if it was ex then this would be elementary.
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  4. #4
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    Sorry i have to say, it look wrong at a glance.
    Why you have to make all these subsitution? make some clear, easy and simple substitution
    for sure your book will give you different answer because your method is incorrect.
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  5. #5
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    How is my method wrong ?


    I'm doing it the right way .

    Okay if i'm wrong then what's the right way ?

    I don't appreciate you telling me i did it wrong if you can't tell me the right thing to do. lol

    u^2 = 1+e^x

    is probably the most clear substitution.
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  6. #6
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    ok, well, u can use it to substitute.
    u^2 = 1+e^x
    2u du=e^x dx
    du/dx=e^x/2u
    integrate sqrt (1+e^x)dx= integrate u.e^x/2u du = 0.5 integrate (u^2-1) du
    what you substitute is incorrect



    sorry i do not know how to type all those notation
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  7. #7
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    perhaps you get confuse about the derivative
    if you integrate f(x) dx
    then you wont to substitute something to that
    so it will become integrate f(u) dx . (du/dx) = f(u) du
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  8. #8
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    Quote Originally Posted by elliotyang View Post
    ok, well, u can use it to substitute.
    u^2 = 1+e^x
    2u du=e^x dx
    du/dx=e^x/2u
    integrate sqrt (1+e^x)dx= integrate u.e^x/2u du = 0.5 integrate (u^2-1) du
    what you substitute is incorrect



    sorry i do not know how to type all those notation

    So that means i'm right

    because

    u^2 = 1+e^x === > u^2 -1 = e^x right ?
    u = sqrt(1 +e^x)


    So what are you talking about ?

    I think you're the confused one lol.

    You don't always have to substitute something that is obvious .
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  9. #9
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    why don't you use my method to compute and see whether my answer is correct or not. since you are not believe what i have done so far.
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  10. #10
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    sorry, just now my mischief cousin playing with my laptop. your method are correct indeed. so sorry if he offended you.
    the one you solve using integration is not encouraged. the working could be quite tedious. but u can free to use any substitution and method u like. as long as the answer is correct.
    Wolfram Mathematica Online Integrator
    this is online integrator, quite useful. you can always check the final answer is correct or not.

    once again, sorry for what my cousin had done to u.
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  11. #11
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    Mathematica sometimes spits out garbage like when i tried to integrate

    \int(tanx)^\frac{1}{2}
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  12. #12
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    sometimes it does happen. perhaps the programming got certain bugs.
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  13. #13
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    Because it isn't a simple integral?

    Let: u^2 = \tan x \ \Leftrightarrow x = \arctan u^2 \ \Rightarrow \ dx = \frac{2u}{1+u^4} du

    So: \int \sqrt{\tan x} \ dx \ = \ \int \frac{2u^2}{1+u^4} \ du

    And then you get some crazy suggestion.
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  14. #14
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    Quote Originally Posted by o_O View Post
    Because it isn't a simple integral?

    Let: u^2 = \tan x \ \Leftrightarrow x = \arctan u^2 \ \Rightarrow \ dx = \frac{2u}{1+u^4} du

    So: \int \sqrt{\tan x} \ dx \ = \ \int \frac{2u^2}{1+u^4} \ du

    And then you get something crazy.

    I already did this question before.

    Adding 2u^2 and -2u^2 to the denominator, makes the problem easy .

    Then after that you can split it into a partial fraction decompose.

    \int \frac{2u^2}{1+u^4} \ du =

    \int \frac{2u^2}{1+2u^2 - 2u^2 +u^4} \ du

    \int \frac{2u^2}{(u^2+1)^2 - 2u^2 } \ du

    using the identity a^2 - b^2 = (a-b)(a+b)

    \int \frac{2u^2}{(u^2 - √2u +1)(u^2 +√2u+1) } \ du
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  15. #15
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    \int \frac{2u^2}{(u^2 - sqrt(2)u +1)(u^2 +sqrt(2)u+1) } \ du
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