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eexiam Group Title

integrate by parts: ∫xln(1+x)dx

  • 2 years ago
  • 2 years ago

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  1. imron07 Group Title
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    Integrating by part will do, i think.

    • 2 years ago
  2. muntoo Group Title
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    \[\large \int u \, dv=uv-\int v \, du.\!\] Set \(u\) to something easily differentiable.

    • 2 years ago
  3. imron07 Group Title
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    u=x dv=ln(1+x)

    • 2 years ago
  4. eexiam Group Title
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    i got |dw:1350949609547:dw|

    • 2 years ago
  5. eexiam Group Title
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    i think its wrong

    • 2 years ago
  6. imron07 Group Title
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    \[u=x, du=dx\\dv=\ln(1+x)dx,v=\frac{1}{1+x}\\\int u\ dv=\frac{x}{1+x}-\int \frac{1}{1+x}dx\]

    • 2 years ago
  7. imron07 Group Title
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    Finally,\[\int u \ dv=\frac{x}{1+x}-\ln(1+x)+C\]

    • 2 years ago
  8. eexiam Group Title
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    ohh ok that makes it easier then.

    • 2 years ago
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