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Babyslapmafro

  • 3 years ago

Please help me get started with solving the following integral (click to see).

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  1. Babyslapmafro
    • 3 years ago
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    \[\int\limits_{}^{}(lnx)^{2}dx\]

  2. Babyslapmafro
    • 3 years ago
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    \[\int\limits_{}^{}u ^{2}\frac{ du }{ 1/x }\] Is this the correct start before doing integration by parts?

  3. OkayOkay
    • 3 years ago
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    u=ln[x]^2, v=x, du=2ln[x]1/x Integration by parts: uv-integral[vdu]

  4. OkayOkay
    • 3 years ago
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    uv-integral[vdu]= xln[x]^2-integral[2xln[x]1/x]= xln[x]^2-2integral[ln[x]]

  5. OkayOkay
    • 3 years ago
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    questions or you got it from here? You will do another integration by parts. So a total of two integration by parts to get to a simple 1dx integral.

  6. OkayOkay
    • 3 years ago
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    step 1: \[xln[x]^{2}-\int\limits_{}^{}2x \ln [x] \frac{ 1 }{ x }\]

  7. OkayOkay
    • 3 years ago
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    simplify: \[xln[x]^2-2\int\limits_{}^{}\ln[x]dx \]

  8. OkayOkay
    • 3 years ago
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    \[xln[x]^2-2(xln[x]-\int\limits_{}^{}x(1/x)dx=xln[x]^2-2(xln[x]-\int\limits_{}^{}1dx\]

  9. OkayOkay
    • 3 years ago
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    \[x(\ln[x]^2-2\ln[x]+2)+c\]

  10. OkayOkay
    • 3 years ago
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    What do you think? I can show you an easy way to remember these multi step problems if you'd like....

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