Using the cover-up method discussed in
http://openstudy.com/study#/updates/500c4d0fe4b0549a89302b3c

Show that
$$
\frac{100}{(x-1) (x+1)^2
\left(x^2+4\right)}=\frac{4
x-4}{x^2+4}+\frac{5}{x-1}-\frac{
9}{x+1}-\frac{10}{(x+1)^2}
$$

Question

Using the cover-up method discussed in
http://openstudy.com/study#/updates/500c4d0fe4b0549a89302b3c

Show that
$$
\frac{100}{(x-1) (x+1)^2
   \left(x^2+4\right)}=\frac{4
   x-4}{x^2+4}+\frac{5}{x-1}-\frac{
   9}{x+1}-\frac{10}{(x+1)^2}
$$

anonymous · Answer

I don't know how to type "fraction" symbol :)

anonymous · Answer

$$\frac{100}{(x-1)(x+1)^2(x^2+4)}= \frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}+\frac{Dx+E}{x^2+4}$$
we will multiply by (x-1)= and let x=1
we will get A=5
by (x+1)^2 let x=-1  C=100/-2(5)=-10
now by x^2+4 and let x=+2i or(-2i) then $$2Di+E=\frac{100}{(-5(2i+1))}=8i-4$$
then D=4 and E=-4
so let x=0 - 25=-A+B+C+E/4
then B=-9 ! 
:) !