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Math Help - Evaluation of Integrals

  1. #1
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    Evaluation of Integrals

    1. Integral of (e^x+1)dx/(e^x-1)

    I tried the method of substitution and i let u be e^x-1. then du=e^xdx so solving for dx=du/e^x. I dont think this is correct. It doesnt make anything easier. Can someone lead me in the right direction?

    2. Integral of t dt/(sqrt(4-t^4))

    I let u = 4-t^4 so du=-4t^3dt, then dt=-du/-4t^3. What do i do now?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Belowzero78 View Post
    1. Integral of (e^x+1)dx/(e^x-1)

    I tried the method of substitution and i let u be e^x-1. then du=e^xdx so solving for dx=du/e^x. I dont think this is correct. It doesnt make anything easier. Can someone lead me in the right direction?
    One of many ways to do this...

    \int \frac {e^x + 1}{e^x - 1}~dx = \int \frac {e^x - 1 + 2}{e^x - 1}~dx

    = \int \left( 1 + \frac 2{e^x - 1} \right)~dx

    = x + \int \frac 2{e^x - 1}~dx

    Now consider \int \frac 2{e^x - 1}~dx

    Let u = e^x - 1 \implies \boxed{e^x = u + 1}

    \Rightarrow du = e^x ~dx

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

    \Rightarrow \frac {du}{u + 1} = dx

    So the integral becomes:

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

    = 2 \int \left( \frac 1u - \frac 1{u + 1} \right)~du

    and I leave you to finish off.

    2. Integral of t dt/(sqrt(4-t^4))

    I let u = 4-t^4 so du=-4t^3dt, then dt=-du/-4t^3. What do i do now?
    Let u = t^2

    \Rightarrow du = 2t~dt \implies \frac 12~du = t~dt

    So your integral becomes:

    \frac 12 \int \frac 1{\sqrt{4 - u^2}}~du

    which is of the form \int \frac 1{\sqrt{a^2 - x^2}}~dx = \sin^{-1} \frac xa + C
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  3. #3
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    Hello, Belowzero78!

    1)\;\;\int\frac{e^x+1}{e^x-1}\,dx
    Here's a back-door approach . . .


    Divide top and bottom by e^{\frac{x}{2}}:

    . . \int \frac{\frac{1}{e^{\frac{x}{2}}}(e^x + 1)} {\frac{1}{e^{\frac{x}{2}}}(e^x-1)} \,dx \;=\;\int\frac{(e^{\frac{x}{2}} + e^{-\frac{x}{2}})\,dx}{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}


    Let: . u \:=\:e^{\frac{x}{2}} - e^{-\frac{x}{2}} \quad\Rightarrow\quad du \:=\:\left(\tfrac{1}{2}e^{\frac{x}{2}} + \tfrac{1}{2}e^{-\frac{x}{2}}\right)\,dx . \Rightarrow \quad\left(e^{\frac{x}{2}} + e^{-\frac{x}{2}}\right)\,dx \:=\:2\,du


    Substitute: . \int\frac{2\,du}{u} \;\;=\;\;2\ln|u| + C \quad\Rightarrow\quad 2\ln\left|e^{\frac{x}{2}} - e^{-\frac{x}{2}}\right| + C

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