University of Florida/Egm4313/S12.team14.savardi

Problem 8

${\displaystyle \displaystyle x^{n}log(1+x)dx}$	${\displaystyle \displaystyle (Eq.1)}$

${\displaystyle \displaystyle n=0,n=1}$	${\displaystyle \displaystyle }$

Find the integral of EQ 1 using integration by parts for n=0 and n=1

For n=0
For n=1

${\displaystyle \displaystyle \int xlog(1+x)dx}$	${\displaystyle \displaystyle }$

Take u=(x+1) and du=dx

${\displaystyle \displaystyle \int (u-1)log(u)du}$	${\displaystyle \displaystyle }$

This gives two separate integrals:

${\displaystyle \displaystyle \int ulog(u)du-\int log(u)du}$	${\displaystyle \displaystyle (Eq.2)}$

Integrate the first integral (left hand) by parts, taking:

${\displaystyle \displaystyle f=log(u),df={\frac {1}{u}}du,dg=udu,g={\frac {u^{2}}{2}}}$	${\displaystyle \displaystyle }$

Giving:

${\displaystyle \displaystyle {\frac {1}{2}}u^{2}log(u)-ulog(u)+\int du-{\frac {1}{2}}\int udu}$	${\displaystyle \displaystyle (Eq.3)}$

Integrate the second integral (right hand) by parts, taking: :

${\displaystyle \displaystyle f=log(u),df={\frac {1}{u}}du,dg=du,g=u}$	${\displaystyle \displaystyle }$

Giving:

${\displaystyle \displaystyle -ulog(u)+\int du+\int ulog(u)du}$	${\displaystyle \displaystyle (Eq.4)}$

Substituting EQ.3 and EQ.4 into EQ.2 and integrating the remaining integrals gives:

${\displaystyle \displaystyle -{\frac {u^{2}}{4}}+{\frac {1}{2}}u^{2}log(u)+u-ulog(u)+Constant(K)}$	${\displaystyle \displaystyle }$

Since u=x+1

${\displaystyle \displaystyle -{\frac {1}{4}}(x+1)^{2}+x+{\frac {1}{2}}(x+1)^{2}log(x+1)-(x+1)log(x+1)+1+K}$	${\displaystyle \displaystyle }$

Giving the final answer of:

${\displaystyle \displaystyle {\frac {1}{4}}(2(x^{2}-1)-(x-2)x+log(x+1))+K}$	${\displaystyle \displaystyle (Eq.5)}$