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Math Help - Integration help

  1. #1
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    Integration help

    Could someone assist on the two problems below?

    1. Find the anti-derivative by parts:

    ln3x/x^2 I get the following answer: -1/x*ln3x-1/2x^2 + C


    2. Solve by substitution:
    x*(x-3)^1/2 If I use u=(x-3)^1/2 I don't know how to get rid of the first x

    Any help will be appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Wildcatz View Post
    Could someone assist on the two problems below?

    1. Find the anti-derivative by parts:

    ln3x/x^2 I get the following answer: -1/x*ln3x-1/2x^2 + C
    this is not correct. try again. what part did you decide to differentiate? which did you integrate?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Wildcatz View Post
    2. Solve by substitution:
    x*(x-3)^1/2 If I use u=(x-3)^1/2 I don't know how to get rid of the first x

    Any help will be appreciated.
    wrong substitution. let u = x - 3
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  4. #4
    Senior Member polymerase's Avatar
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    Let u=ln(3x) and dv=x^{-2}\:dx

    Thus, du=\dfrac{1}{x}\:dx and v=-\dfrac{1}{x}

    \int \dfrac{ln(3x)}{x^2} = -\dfrac{ln(3x)}{x}+\int\dfrac{1}{x^2}
    \int \dfrac{ln(3x)}{x^2} = -\dfrac{ln(3x)}{x} + ({-\dfrac{1}{x}})

    Therefore, \int \dfrac{ln(3x)}{x^2} = -(\dfrac{ln(3x)+1}{x})
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  5. #5
    Super Member PaulRS's Avatar
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    \int{x\cdot{\sqrt[]{x-3}}}dx=2\cdot{\int{\frac{x\cdot{(x-3)}}{2\cdot{\sqrt[]{x-3}}}}dx}

    Let u=\sqrt[]{x-3} thus: \frac{du}{dx}=\frac{1}{2\cdot{\sqrt[]{x-3}}}

    Then: \int{x\cdot{\sqrt[]{x-3}}}dx=2\cdot{\int{(u^2+3)\cdot{u^2}}du}

    2\cdot{\int{(u^2+3)\cdot{u^2}}du}=2\cdot{\left(\fr  ac{u^5}{5}+u^3\right)}+k

    So \int{x\cdot{\sqrt[]{x-3}}}dx=2\cdot{\left(\frac{(\sqrt[]{x-3})^5}{5}+(\sqrt[]{x-3})^3\right)}+k

    Which is: \int{x\cdot{\sqrt[]{x-3}}}dx=\frac{2}{5}\cdot{(x-3)^{3/2}\cdot{(x+2)}}+k
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Wildcatz View Post
    2. Solve by substitution:
    x*(x-3)^1/2 If I use u=(x-3)^1/2 I don't know how to get rid of the first x

    Quote Originally Posted by Jhevon View Post
    wrong substitution. let u = x - 3
    I think it's quickly substitution, this leads to (by setting u^2=x-3),

    \int {x\sqrt {x - 3} \,dx} = 2\int {u^2 \left( {u^2 + 3} \right)\,du} .

    The conclusion follows from PaulRS' evaluation.
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