How do you use partial fraction decomposition to decompose the fraction to integrate (6s^3) /(s^(4) - 5s^(2) +4)6s3s45s2+4?

1 Answer
May 22, 2015

The denominator can be factored as:

s^4-5s^2+4=(s^2-4)(s^2-1)=(s-2)(s+2)(s-1)(s+1)s45s2+4=(s24)(s21)=(s2)(s+2)(s1)(s+1)

Hence, we can write (6s^3)/(s^4-5s^2+4)=A/(s-2)+B/(s+2)+C/(s-1)+D/(s+1)6s3s45s2+4=As2+Bs+2+Cs1+Ds+1

Now multiply everything by (s-2)(s+2)(s-1)(s+1)(s2)(s+2)(s1)(s+1) to get

6s^3=A(s+2)(s-1)(s+1)+B(s-2)(s-1)(s+1)+C(s-2)(s+2)(s+1)+D(s-2)(s+2)(s-1)6s3=A(s+2)(s1)(s+1)+B(s2)(s1)(s+1)+C(s2)(s+2)(s+1)+D(s2)(s+2)(s1)

You want this equation to be true for all ss. This allows you to solve for A, B, CA,B,C, and DD. The quickest way to do this is to plug in, successively, s=1s=1, s=-1s=1, s=2s=2, and s=-2s=2 (even though the original equation was undefined at these values).

s=1\Rightarrow 6=C\cdot (-1)\cdot 3\cdot 2=-6C\Rightarrow C=-1s=16=C(1)32=6CC=1

s=-1\Rightarrow -6=D\cdot (-3)\cdot 1\cdot (-2)=6D\Rightarrow D=-1s=16=D(3)1(2)=6DD=1

s=2\Rightarrow 48=A\cdot 4\cdot 1\cdot 3=12A\Rightarrow A=4s=248=A413=12AA=4

s=-2\Rightarrow -48=B\cdot (-4)\cdot (-3)\cdot (-1)=-12B\Rightarrow B=4s=248=B(4)(3)(1)=12BB=4

Hence,

\int\frac{6s^2}{s^4-5s^2+4}\ ds

=\int\frac{4}{s-2}\ ds+\int\frac{4}{s+2}\ ds-\int\frac{1}{s-1}\ ds-\int\frac{1}{s+1}\ ds

=4ln|s-2|+4ln|s+2|-ln|s-1|-ln|s+1|+C

=ln|\frac{(s^{2}-4)^{4}}{s^2-1}|+C