Question

Evaluate the integral
x4 – 2x2 + 4x + 1
x3 – x2 - x + 1
dx

Answer 1

Looks like the integral is

`\displaystyle\int(x^4-2x^2+4x+1)/(x^3-x^2-x+1)\,\mathrm dx`

First simplify the integrand by long division (or however you like):

`(x^4-2x^2+4x+1)/(x^3-x^2-x+1)=x+1+(4x)/(x^3-x^2-x+1)`

Also notice that

`x^3-x^2-x+1=x^2(x-1)-(x-1)=(x^2-1)(x-1)=(x+1)(x-1)^2`

Split the last term into partial fractions:

`(4x)/((x+1)(x-1)^2)=\frac a{x+1}+\frac b{x-1}+\frac c{(x-1)^2}`

`4x=a(x-1)^2+b(x+1)(x-1)+c(x+1)`

`4x=(a+b)x^2+(c-2a)x+a-b+c`

`\implies\begin{cases}a+b=0\\c-2a=4\\a-b+c=0\end{cases}\implies a=-1,b=1,c=2`

So the integral is equivalent to

`\displaystyle\int\left(x+1-\frac1{x+1}+\frac1{x-1}+\frac2{(x-1)^2}\right)\,\mathrm dx`

`=\displaystyle\frac{x^2}2+x-\ln|x+1|-\ln|x-1|-\frac2{x-1}+C`

`=\displaystyle\frac{x^2}2+x-\ln|x^2-1|-\frac2{x-1}+C`