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eexiam

  • 3 years ago

integrate by parts: ∫xln(1+x)dx

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  1. imron07
    • 3 years ago
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    Integrating by part will do, i think.

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

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

  4. eexiam
    • 3 years ago
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    i got |dw:1350949609547:dw|

  5. eexiam
    • 3 years ago
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    i think its wrong

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

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

  8. eexiam
    • 3 years ago
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    ohh ok that makes it easier then.

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