Errors in partial fractions decomposition I decompose the fraction $\frac{x^2}{(x^2+1)^3}$ into $$\frac{x^2}{(x^2+1)^3}=\frac{Ax+B}{(x^2+1)^2}+\frac{Cx+D}{(x^2+1)^3}.$$
From here I found: $A=C=0,$ and $B=-D=1$
Now $$\frac{x^2}{(x^2+1)^3}=\frac{1}{(x^2+1)^2}-\frac{1}{(x^2+1)^3}$$
but the right side is not equal to the left side. How is this possible.
And, and similarly another example How is this possible
$$\frac{x^5+1}{x^4(x^2+1)}=\frac{Ax^3+Bx^2+Cx+D}{x^4}+\frac{Ex+F}{x^2+1}.$$
Find, $A=B=C=F=0,$ and $D=E=1$
Now write, $$\frac{x^5+1}{x^4(x^2+1)}=\frac{1}{x^4}+\frac{1}{x^2+1}.$$
but the right side is not equal to the left side. How is this possible.
Please help me. Thanks.

The first expression you got it right, I do the second for you:
$\dfrac{x^5+1}{x^4(x^2+1)}=\dfrac{x^5}{x^4(x^2+1)}+\dfrac{1}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1+x^2-x^2}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1+x^2}{x^4(x^2+1)}-\dfrac{x^2}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1}{x^4}-\dfrac{1}{x^2(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1}{x^4}-\dfrac{1}{x^2}+\dfrac{1}{x^2+1}$
