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Sorry, wrong question! the integral of (x(ln(1+x))dx
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integration by parts
xdx=(x^2)/2+c
?
have you done integration by parts before?
Yes, but I cannot figure this one out. What shoul u be? ln(1+x)?
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\[u=\ln(1+x)\] \[dv=xdx\]
Used that, but just keeps growing.
how ...
\[du=\frac{1}{1+x}dx\] \[v=\frac{x^2}{2}\] \[\int x\ln(1+x)dx=\frac{x^2}{2}\ln(1+x)-\int\frac{x^2}{2(1+x)}dx\]
right?
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a little algebra gives us \[\frac{x^2}{2(1+x)}=\frac{1}{2(1+x)}+\frac{x}{2}-\frac{1}{2}\]
partial fractions?
long division
you don't use partial fractions when the power if the numerator is >= power of the denominator
thanks got it now
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