University of Florida/Egm4313/S12.team14.savardi

Given

$x^{n}log(1+x)dx$

$(Eq.1)$

$n=0,n=1$

${\displaystyle }$

Find

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

Solution

For n=0
For n=1

$\int xlog(1+x)dx$

${\displaystyle }$

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

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

${\displaystyle }$

This gives two separate integrals:

$\int ulog(u)du-\int log(u)du$

$(Eq.2)$

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

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

${\displaystyle }$

Giving:

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

$(Eq.3)$

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

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

${\displaystyle }$

Giving:

$-ulog(u)+\int du+\int ulog(u)du$

$(Eq.4)$

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

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

${\displaystyle }$

Since u=x+1

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

${\displaystyle }$

Giving the final answer of:

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

$(Eq.5)$