# University of Florida/Egm4313/S12.team14.savardi

Problem 8

## Given

 $\displaystyle x^{n}log(1+x)dx$ $\displaystyle (Eq.1)$ $\displaystyle n=0,n=1$ $\displaystyle \displaystyle$

## Find

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

## Solution

For n=0
For n=1

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

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

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

This gives two separate integrals:

 $\displaystyle \int ulog(u)du-\int log(u)du$ $\displaystyle (Eq.2)$ Integrate the first integral (left hand) by parts, taking:

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

Giving:

 $\displaystyle {\frac {1}{2}}u^{2}log(u)-ulog(u)+\int du-{\frac {1}{2}}\int udu$ $\displaystyle (Eq.3)$ Integrate the second integral (right hand) by parts, taking: :

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

Giving:

 $\displaystyle -ulog(u)+\int du+\int ulog(u)du$ $\displaystyle (Eq.4)$ Substituting EQ.3 and EQ.4 into EQ.2 and integrating the remaining integrals gives:

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

Since u=x+1

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

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