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)#
Hence, we can write #(6s^3)/(s^4-5s^2+4)=A/(s-2)+B/(s+2)+C/(s-1)+D/(s+1)#
Now multiply everything by #(s-2)(s+2)(s-1)(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)#
You want this equation to be true for all #s#. This allows you to solve for #A, B, C#, and #D#. The quickest way to do this is to plug in, successively, #s=1#, #s=-1#, #s=2#, and #s=-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=-1#
#s=-1\Rightarrow -6=D\cdot (-3)\cdot 1\cdot (-2)=6D\Rightarrow D=-1#
#s=2\Rightarrow 48=A\cdot 4\cdot 1\cdot 3=12A\Rightarrow A=4#
#s=-2\Rightarrow -48=B\cdot (-4)\cdot (-3)\cdot (-1)=-12B\Rightarrow B=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#
