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Math Help - Simple Integration

  1. #1
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    Simple Integration

    \int \frac{u}{\sqrt{u^2 +16}} \ du

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Air View Post
    \int \frac{u}{\sqrt{u^2 +16}} \ du

    Thanks in advance.
    Make the substitution w = u^2 + 16. Note that \frac{dw}{du} = 2u \Rightarrow du = \frac{dw}{2u} ....
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  3. #3
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    Hi

    \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)=?
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)=?
    \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right) = \frac{ \mathrm{d}}{\mathrm{d}u} \left( u^2+16 \right)^\frac{1}{2} = \frac{1}{2}(u^2 + 16)^{-\frac{1}{2}}(2u) = (u)(u^2 + 16)^{-\frac{1}{2}} = \frac{u}{\sqrt{u^2+16}}
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Make the substitution w = u^2 + 16. Note that \frac{dw}{du} = 2u \Rightarrow du = \frac{dw}{2u} ....
    You should try this way to see why flyingsquirrel was able to suggest you consider the derivative \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right) ......
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    You should try this way to see why flyingsquirrel was able to suggest you consider the derivative \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right) ......
    \int \frac{u}{\sqrt{u^2 + 16}} \ du

    w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}

    \int \frac{u}{\sqrt w} . \frac{dw}{2u}

    \int \frac{1}{\sqrt w} . \frac{dw}{2}

    \frac{1}{2} \int \frac{1}{\sqrt w} \ dw

    \frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right]

    w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}
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  7. #7
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    Quote Originally Posted by Air View Post
    \int \frac{u}{\sqrt{u^2 + 16}} \ du

    w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}

    \int \frac{u}{\sqrt w} . \frac{dw}{2u}

    \int \frac{1}{\sqrt w} . \frac{dw}{2}

    \frac{1}{2} \int \frac{1}{\sqrt w} \ dw

    \frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right]

    w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}
    + C.

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