University of Florida/Egm4313/S12.team14.savardi

Given

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x^{n}log(1+x)dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
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n=0, n=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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Find
Find the integral of EQ 1 using integration by parts for n=0 and n=1

Solution
For n=0 For n=1


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\int xlog(1+x)dx$$ $$ Take u=(x+1) and du=dx
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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\int (u-1)log(u)du$$ $$ This gives two separate integrals:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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\int u log(u)du-\int log(u)du$$ $$ Integrate the first integral (left hand) by parts, taking:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
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f=log(u), df=\frac{1}{u}du, dg=udu, g=\frac{u^{2}}{2}$$ $$ Giving:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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\frac{1}{2}u^{2}log(u)-ulog(u)+\int du-\frac{1}{2}\int udu$$ $$ Integrate the second integral (right hand) by parts, taking: :
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
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f=log(u), df=\frac{1}{u}du, dg=du, g=u$$ $$ Giving:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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-ulog(u)+\int du+\int ulog(u)du$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
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Substituting EQ.3 and EQ.4 into EQ.2 and integrating the remaining integrals gives:
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-\frac{u^{2}}{4}+\frac{1}{2}u^{2}log(u)+u-ulog(u)+ Constant (K) $$ $$ Since u=x+1
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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-\frac{1}{4}(x+1)^{2}+x+\frac{1}{2}(x+1)^{2}log(x+1)-(x+1)log(x+1)+ 1 + K$$ $$ Giving the final answer of:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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\frac{1}{4}(2(x^{2}-1)-(x-2)x+log(x+1)) + K $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq.5)
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