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Math Help - rational substiution integral

  1. #1
    Member akhayoon's Avatar
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    rational substiution integral

     \int \frac{dx}{\sqrt{1+e^{x}}}


     u^{2}=1+e^{x}

    [tex] 2u du=e^{x} dx [tex]

    I finally get to

     \frac{2}{3} (\int \frac{1}{u-1}du+ \int \frac{u+1}{u^{2}+u+1}du)


    so if this is right

    can somebody help me with hte second part of the last integral

    and if it's wrong somekind of solution
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by akhayoon View Post
     \int \frac{dx}{\sqrt{1+e^{x}}}


     u^{2}=1+e^{x}

    [tex] 2u du=e^{x} dx [tex]

    I finally get to

     \frac{2}{3} (\int \frac{1}{u-1}du+ \int \frac{u+1}{u^{2}+u+1}du)


    so if this is right

    can somebody help me with hte second part of the last integral

    and if it's wrong somekind of solution
    i get something different for the integral. by the way, i don't like the substitution you made because \sqrt{u^2} = |u| \ne u, but anyway, you'd get something like 2 \int \frac {du}{u^2 - 1} if we say u > 0

    an alternative is to use two substitutions. first u = 1 + e^x and then t = \sqrt{u} and you'll get a similar integral to the one above

    if you're still interested, completing the square coupled with substitution can take care of your original integral
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  3. #3
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    Hello, akhayoon!

    Your substitution should have come out like Jhevon's . . .


     \int \frac{dx}{\sqrt{1+e^{x}}}

    Let u = \sqrt{1+e^x}\quad\Rightarrow\quad e^x \:=\:u^2-1\quad\Rightarrow\quad x \:=\:\ln(u^2-1) \quad\Rightarrow\quad  dx \:=\:\frac{2u\,du}{u^2-1}

    Substitute: . \int\frac{1}{u}\cdot\frac{2u\,du}{u^2-1} \;=\;2\int\frac{du}{u^2-1} \;=\;\ln\left|\frac{u-1}{u+1}\right| + C

    Back-substitute: . \boxed{\ln\left|\frac{\sqrt{1+e^x} - 1}{\sqrt{1+e^x} + 1}\right| + C}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Here's another method . . . it's either brilliant or incredibly stupid.

    Let e^x \:=\:\tan^2\theta\quad\Rightarrow\quad x \:=\:2\ln|\tan\theta|\quad\Rightarrow\quad dx \:=\:\frac{2\sec^2\!\theta}{\tan\theta}\,d\theta

    Substitute: . \int\frac{1}{\sec\theta}\cdot\frac{2\sec^2\!\theta  }{\tan\theta}\,d\theta \;=\;2\int\csc\theta\,d\theta \;=\;2\!\cdot\!\ln|\csc\theta - \cot\theta| + C


    Back-substitute

    We have: . \tan\theta \:=\:e^{\frac{x}{2}}

    This is a right triangle with: .  opp = e^{\frac{x}{2}},\;adj = 1
    Hence: . hyp = \sqrt{1+e^x} \quad\Rightarrow\quad\csc\theta = \frac{\sqrt{1+e^x}}{e^{\frac{x}{2}}},\;\cot\theta = \frac{1}{e^{\frac{x}{2}}}


    So we have: . 2\!\cdot\!\ln\left|\frac{\sqrt{1+e^x}}{e^{\frac{x}  {2}}} - \frac{1}{e^{\frac{x}{2}}}\right| + C \;=\;2\!\cdot\!\ln\left|\frac{\sqrt{1+e^x} - 1}{e^{\frac{x}{2}}}\right| + C

    . . = \;2\ln\left|\sqrt{1+e^x} - 1\right| - 2\ln\left|e^{\frac{x}{2}}\right| + C \;=\;2\!\cdot\!\ln\left|\sqrt{1+e^x} - 1\right| - \ln\left(e^{\frac{x}{2}}\right)^2 + C

    . . = \;2\!\cdot\!\ln\left|\sqrt{1+e^x} - 1\right| - \ln(e^x) + C \;=\;\boxed{2\!\cdot\!\ln\left|\sqrt{1+e^x} - 1\right| - x + C}<br />



    You mission (should you decide to accept it)
    . . is to prove that my two answers are equivalent.
    .
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Here's another method . . . it's either brilliant or incredibly stupid.
    i say brilliant! so crazy, it's brilliant! ...though not for the faint of heart, or those new to the world of integrals. how did you discover such a substitution? was the integral given in a "special form"?

    You mission (should you decide to accept it)
    . . is to prove that my two answers are equivalent.
    .
    later
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  5. #5
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    Krizalid's Avatar
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    One can play with trig. sub. as many times as necessary.

    700% personal opinion, I don't like trig. substitutions, some times it turns the integrand into a mess, but it depends of the problem.

    I prefer workin' with algebra, you can get faster solutions, especially the reciprocal substitution, which can turn a nasty integral, into a cute one. (Not always, of course.)
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