Is it possible to $\int\frac{x^2 - 4 }{(x^2 + 1)(x^2 + 2)(x^2 + 3)} \, dx$ without partial fractions

Question

$$\int\frac{x^2 - 4 }{(x^2 + 1)(x^2 + 2)(x^2 + 3)} \, dx$$ I did it by partial fractions, I am not typing it here because it is very long, if you want I can.

I want to do it without partial fractions.

Answer 1

$$\frac{x^2 - 4 }{(x^2 + 1)(x^2 + 2)(x^2 + 3)}=\frac{1.5}{(x^2 + 1)(x^2 + 2)}-\frac{0.5}{(x^2 + 2)(x^2 + 3)}$$ and $$\frac{1.5}{(x^2 + 1)(x^2 + 2)}=\frac{1.5}{(x^2 + 1)}-\frac{1.5}{(x^2 + 2)}$$ $$\frac{0.5}{(x^2 + 2)(x^2 + 3)}=\frac{0.5}{(x^2 + 2)}-\frac{0.5}{(x^2 + 3)}$$

Answer 2

Partial Fraction Decomposition won't be that painful if we choose $x^2=y$

$$\dfrac{y-4}{(y+1)(y+2)(y+3)}=\dfrac A{y+1}+\dfrac B{y+2}+\dfrac C{y+3}$$

Method$\#1:$ $\implies y-4=A(y+3)(y+2)+\cdots$

Set $y=-1,-2,-3$ one by one to find $A,B,C$

Method$\#2:$

$y-4=y^2(A+B+C)+y(5A+4B+3C)+(6A+3B+2C)$ Can you find $A,B,C$ using the facts $$A+B+C=0,5A+4B+3C=1,6A+3B+2C=-4$$

Answer 3

Just thought I'd point out the following just in case you've come across this kind of stuff.

If the you instead wanted the value of the integral evaluated over some interval, you could extend it to the complex plane (using something like a semi-circle or rectangle) and then use the residue theorem. Although this method would avoid partial fractions there would probably be just as much algebra needed to manipulate complex numbers!