Question

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Answer 1

`(x^2+x+1)/((x+1)^2(x+2))=(a_1x+a_0)/((x+1)^2)+\frac b{x+2}`

`\implies(x^2+x+1)/((x+1)^2(x+2))=((a_1x+a_0)(x+2)+b(x+1)^2)/((x+1)^2(x+2))`

`\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)`

`\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3`

So you have


`\displaystyle\int_0^2(x^2+x+1)/((x+1)^2(x+2))\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2(\mathrm dx)/((x+1)^2)+3\int_0^2(\mathrm dx)/(x+2)`

`=\displaystyle-2\int_1^3(x-1)/(x^2)\,\mathrm dx-\int_0^2(\mathrm dx)/((x+1)^2)+3\int_0^2(\mathrm dx)/(x+2)`

where in the first integral we substitute
`x\mapsto x+1`.


`=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_(x=0)^(x=2)+3\ln|x+2|\bigg|_(x=0)^(x=2)`

`=-2\left(\ln|x|+\frac1x\right)\bigg|_(x=1)^(x=3)-\frac23+3(\ln4-\ln2)`

`=-2\left(\ln3+\frac13-1\right)-\frac23+3\ln2`

`=\frac23+\ln\frac89`