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Math Help - 2 more intergrals

  1. #1
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    2 more intergrals

    Last two intergrals and I'm done for today's homework.

    1) 3e^x / sqrt(1-e^(2x))

    2) (3-x) / (x^2+4)

    Also I got one intergral down to 1/u du, and is 1/u the same as ln|u|??
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Hibijibi View Post
    1) 3e^x / sqrt(1-e^(2x))
    You can either substitute t=\exp x or remember that \frac{\mathrm{d}}{\mathrm{d}x}(\arcsin u(x))=\frac{u'(x)}{\sqrt{1-u^2(x)}}
    2) (3-x) / (x^2+4)
    You may rewrite it as \frac{3-x}{x^2+4}=\frac{3}{x^2+4}-\frac{x}{x^2+4}<br />
=\frac{3}{4}\frac{1}{1+\left(\frac{x}{2}\right)^2}-\frac{1}{2}\frac{2x}{x^2+4}
    Also I got one intergral down to 1/u du, and is 1/u the same as ln|u|??
    No, that's wrong. (except if u(x)=x )
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  3. #3
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    I solved the first one, but the second still confused me a bit on how to approach it. Should I split into two separate integrals? And what should u be for those?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Hibijibi View Post
    Last two intergrals and I'm done for today's homework.

    1) 3e^x / sqrt(1-e^(2x))

    2) (3-x) / (x^2+4)

    Also I got one intergral down to 1/u du, and is 1/u the same as ln|u|??
    1) \int\frac{3e^x}{\sqrt{1-e^{2x}}}\,dx

    You can rewrite the integral as follows:

    \int\frac{3e^x}{\sqrt{1-(e^x)^2}}\,dx

    Now our substitution will become evident. Let u=e^x. du=e^x dx\rightarrow dx=\frac{du}{e^x}.

    Substitution into the integral, we have:

    \int\frac{3e^x}{\sqrt{1-u^2}}\frac{du}{e^x}=3\int\frac{du}{\sqrt{1-u^2}}

    The integral of this is sin^{-1}u+C. Substitute the u value back in, and you get:

    \int\frac{3e^x}{\sqrt{1-e^{2x}}}\,dx=3sin^{-1}{\left(e^x\right)}+C

    2) \int\frac{3-x}{x^2+4}\,dx

    I would recommend rewriting the integrand and then break it up into 2 integrals.

    \int\frac{3-x}{x^2+4}\,dx=\int\frac{3}{x^2+4}\,dx-\int\frac{x}{x^2+4}\,dx.

    The first integral will take on the form of tan^{-1}x, in particular, it will be \frac{1}{2}tan^{-1}{\left(\frac{x}{2}\right)}+C.

    The second integral requires a substitution: u=x^2+4. du=2x\rightarrow dx=\frac{du}{2x}. Substituting into the second integral, we have:

    \int\frac{x}{u}\frac{du}{2x}=\frac{1}{2}\int\frac{  du}{u}.

    This integral wll be ln{\left|u\right|}+C. Substitute u back in and you get ln{\left|x^2+4\right|}+C.

    Thus:

    \int\frac{3}{x^2+4}\,dx-\int\frac{x}{x^2+4}\,dx=\frac{3}{2}tan^{-1}{\left(\frac{x}{2}\right)}-ln{\left|x^2+4\right|}+C

    Hope this answers your question!!!
    Last edited by Chris L T521; May 8th 2008 at 09:21 AM.
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  5. #5
    Super Member flyingsquirrel's Avatar
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    I solved the first one, but the second still confused me a bit on how to approach it. Should I split into two separate integrals? And what should u be for those?
    Yes, splitting it into two integrals is a good idea :

    \int \frac{3-x}{x^2+4}\,\mathrm{d}x=\frac{3}{2}\int \frac{\frac{1}{2}}{1+\left(\frac{x}{2}\right)^2}\,  \mathrm{d}x-\frac{1}{2}\int\frac{2x}{x^2+4}\,\mathrm{d}x

    For the first one, you can try X=\frac{x}{2} and for the second one, u=x^2+4. Otherwise, I wrote them so that you can integrate directly if you "see" the derivatives which are involved.
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  6. #6
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    Quote Originally Posted by Hibijibi View Post
    I solved the first one, but the second still confused me a bit on how to approach it. Should I split into two separate integrals? And what should u be for those?
    \int \frac{3-x}{x^2+4}\,\mathrm{d}x

    = \int \frac{3}{x^2+4}\,\mathrm{d}x - \int \frac{x} {x^2+4}\,\mathrm{d}x

    = 3 \int \frac{1}{x^2+4}\,\mathrm{d}x - \int \frac{x} {x^2+4}\,\mathrm{d}x

    First integration is just a standard integration ( \mathrm{arctan}) and the second one, you could use substitution with u=x^2 + 4.
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