## Babyslapmafro Group Title Please help me get started with solving the following integral (click to see). one year ago one year ago

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$\int\limits_{}^{}(lnx)^{2}dx$

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$\int\limits_{}^{}u ^{2}\frac{ du }{ 1/x }$ Is this the correct start before doing integration by parts?

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u=ln[x]^2, v=x, du=2ln[x]1/x Integration by parts: uv-integral[vdu]

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uv-integral[vdu]= xln[x]^2-integral[2xln[x]1/x]= xln[x]^2-2integral[ln[x]]

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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.

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step 1: $xln[x]^{2}-\int\limits_{}^{}2x \ln [x] \frac{ 1 }{ x }$

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simplify: $xln[x]^2-2\int\limits_{}^{}\ln[x]dx$

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$xln[x]^2-2(xln[x]-\int\limits_{}^{}x(1/x)dx=xln[x]^2-2(xln[x]-\int\limits_{}^{}1dx$

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$x(\ln[x]^2-2\ln[x]+2)+c$

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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....