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Math Help - Indefinite integration with u-substitution

  1. #1
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    Indefinite integration with u-substitution

    There are three problems where I had difficulty answering:
    the first is (sqrt) ((x-1)/(x^5))

    The first thing i did was distribute the x^5 and got:

    (sqrt)(x^(-4) - x^(-5))...

    I gave the value (x^(-4) - x^(-5)) for u and got:

    du = -4x^(-3) + 5x^(-4). I am not sure if I am doing this right so far.

    THe answer is (2/3)(1-(1/x))^(3/2) + C

    The second is dz/(1 + e^z)

    I used (1 + e^z) to equal u and du = e^z.

    This is where I got confused.

    the answer was: z - ln(1+e^z) + C... the z part baffled me more.
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  2. #2
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    Tehran
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    Wink

    The Second Integral :

    <br />
\int \frac{dz}{1+(e^z)}= \int \frac{(e^z)dz}{(1+(e^z))e^z} \rightarrow u=e^z \rightarrow du = dz e^z \rightarrow \int \frac{du}{(u+1)(u)}<br />
    <br />
\rightarrow \int \frac{du}{u} - \int\frac{du}{u+1} = ln u - ln (u+1) =ln e^z - ln ((e^z)+1) = z - ln((e^z)+1)<br />
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  3. #3
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    The Second Integral :

    <br />
\int \frac{dz}{1+(e^z)}= \int \frac{(e^z)dz}{(1+(e^z))e^z} \rightarrow u=e^z \rightarrow du = dz e^z \rightarrow \int \frac{du}{(u+1)(u)}<br />

    Now we are going to seprate

    <br />
\rightarrow \int \frac{du}{u} - \int\frac{du}{u+1} = ln u - ln (u+1) =z - ln ((e^z)+1)<br />
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