anonymous
  • anonymous
integrate by parts: ∫xln(1+x)dx
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Integrating by part will do, i think.
anonymous
  • anonymous
\[\large \int u \, dv=uv-\int v \, du.\!\] Set \(u\) to something easily differentiable.
anonymous
  • anonymous
u=x dv=ln(1+x)

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anonymous
  • anonymous
i got |dw:1350949609547:dw|
anonymous
  • anonymous
i think its wrong
anonymous
  • anonymous
\[u=x, du=dx\\dv=\ln(1+x)dx,v=\frac{1}{1+x}\\\int u\ dv=\frac{x}{1+x}-\int \frac{1}{1+x}dx\]
anonymous
  • anonymous
Finally,\[\int u \ dv=\frac{x}{1+x}-\ln(1+x)+C\]
anonymous
  • anonymous
ohh ok that makes it easier then.

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