Question

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

Answer 1

Looks like the integral is

$\displaystyle\int\frac{x^4-2x^2+4x+1}{x^3-x^2-x+1}\,\mathrm dx$

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

$\dfrac{x^4-2x^2+4x+1}{x^3-x^2-x+1}=x+1+\dfrac{4x}{x^3-x^2-x+1}$

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:

$\dfrac{4x}{(x+1)(x-1)^2}=\dfrac a{x+1}+\dfrac b{x-1}+\dfrac c{(x-1)^2}$

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

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

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

So the integral is equivalent to

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

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

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