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Math Help - [SOLVED] Indefinite Integration by Sub

  1. #1
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    [SOLVED] Indefinite Integration by Sub

    Integrate:

    a) \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx<br />

    Let\, u = x^2 + 1, du = 2x\, dx

    = \int \frac{2xln(x^2 + 1)}{u} \frac{1}{2x}\, du

    = \int ln(x^2 + 1) u^{-1}\, du

    = ln(x^2 + 1) (ln(x^2 + 1)) + c

    = [ln(x^2 + 1)]^2 + c

    answer is \frac{1}{2} [ln(x^2 + 1)]^2 + c, where does 1/2 come from??


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

    Let \, u = 2x + 1\, du = 2\, dx

    = \frac{1}{2} \int x(u^{-1})\, du

    = (\frac{1}{2})(\frac{1}{2})(x^2)(ln(2x + 1)) + c

    = \frac{1}{4}x^2 (ln|2x + 1|) + c

    answer is \frac{1}{2}x - \frac{1}{4}(ln|2x + 1|) + c, what did I do wrong?
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  2. #2
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    Quote Originally Posted by Macleef View Post
    Integrate:

    a) \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx<br />

    Let\, u = x^2 + 1, du = 2x\, dx

    = \int \frac{2xln(x^2 + 1)}{u} \frac{1}{2x}\, du

    = \int ln(x^2 + 1) u^{-1}\, du

    = ln(x^2 + 1) (ln(x^2 + 1)) + c

    = [ln(x^2 + 1)]^2 + c

    answer is \frac{1}{2} [ln(x^2 + 1)]^2 + c, where does 1/2 come from??


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

    Let \, u = 2x + 1\, du = 2\, dx

    = \frac{1}{2} \int x(u^{-1})\, du

    = (\frac{1}{2})(\frac{1}{2})(x^2)(ln(2x + 1)) + c

    = \frac{1}{4}x^2 (ln|2x + 1|) + c

    answer is \frac{1}{2}x - \frac{1}{4}(ln|2x + 1|) + c, what did I do wrong?
    When you're putting substitutions on integral you're not subst. everything, you're working with both x and u. When you substitute something with u, you have to express everthing inside of integral with u.
    In that case in a) after sub. the integral would be. \int{\frac{\ln{u} du}{u}}. Here substitute again \ln{u}=t \Rightarrow \frac{du}{u}=dt
    so the integral will be \int{t dt}= \frac{t^2}{2}=\frac{ln^2{u}}{2}= \frac{(\ln{x+1})^2}{2}+c
    The same mistake in part b).
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  3. #3
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    Quote Originally Posted by javax View Post
    When you're putting substitutions on integral you're not subst. everything, you're working with both x and u. When you substitute something with u, you have to express everthing inside of integral with u.
    In that case in a) after sub. the integral would be. \int{\frac{\ln{u} du}{u}}. Here substitute again \ln{u}=t \Rightarrow \frac{du}{u}=dt
    so the integral will be \int{t dt}= \frac{t^2}{2}=\frac{ln^2{u}}{2}= \frac{(\ln{x+1})^2}{2}+c
    The same mistake in part b).
    I don't understand where "t" comes from... why is lnu = t?

    also i followed your advice to turn every x variable to u for the second question and still not getting it...

    \int \frac{u}{2u + 1} du<br />

    \int 2u^2 + u du<br />


    \frac{2}{3}(x)^3 + \frac{1}{2}(x )^2 + c<br />
    Last edited by Macleef; February 11th 2009 at 12:20 PM.
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  4. #4
    Member javax's Avatar
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    Because we want to solve it by substitution...That is the correct substitution because when you differentiate \ln{u}=t it becomes

    \frac{du}{u}=dt so now the expression  \frac{du}{u} under integral you'll sub. with dt. When you finish the integral you have to get back the substitutions.

    In part b). after doing those substitutions you'll have

    2x+1 = u

    x=\frac{u-1}{2}

    dx=\frac{du}{2}

    now the integral will be

    \int{\frac{\frac{u-1}{2}}{u} \frac{1}{2}du}
    <br />
=\frac{1}{4}\int{\frac{u-1}{u}du} = \frac{1}{4}\int{du}-\frac{1}{4}\int{\frac{du}{u}} = ... now I believe you can finish it.
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  5. #5
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    a) \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx<br />

    note ...

    \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx = \int \ln(x^2+1) \cdot \frac{2x}{x^2+1} \, dx

    u = \ln(x^2 + 1)

    du = \frac{2x}{x^2+1} \, dx

    \int u \, du = \frac{u^2}{2} + C = \frac{1}{2}[\ln(x^2+1)]^2 + C



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

    u = 2x+1

    x = \frac{u-1}{2}

    du = 2\, dx

    \frac{1}{2}\int \frac{u-1}{2u} \, du

    \frac{1}{4} \int 1 - \frac{1}{u} \, du

    \frac{1}{4}(u - \ln|u|) + C

    \frac{1}{4}[(2x+1) - \ln|2x+1|] + C

    \frac{x}{2} + \frac{1}{4} - \frac{1}{4}\ln|2x+1| + C

    \frac{x}{2} - \frac{1}{4}\ln|2x+1| + C
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