<s>[INST] Partial Fraction Decomposition of $\frac{1}{x^2(x^2+25)}$ I have been reviewing some integration techniques and have been searching for tough integrals with solutions online. When I was going through the solution, however, I found a discrepancy between my solution and theirs and think what I did was correct instead.
I am trying to solve the indefinite integral: $\int\frac{dx}{x^2(x^2+25)}$. My first step was to break it into the fractions $$\frac{1}{x^2(x^2+25)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+25}$$ Then multiplying both sides by $x^2(x^2+25)$, we find our basic equation to be$$1=A*x(x^2+25)+B*(x^2+25)+(Cx+D)*x^2$$ Solving the system of linear equations, I found that $B=\frac{1}{25}$, $D=\frac{-1}{25}$, and $A=C=0$.
This is where I found the discepancy. The online solution has the basic equation as $$1=A*x(x^2+25)+B*(x^2+25)+(Cx+D)*x$$ so when they solve for coefficients they find that $B=\frac{1}{25}$, $C=\frac{-1}{25}$, and $A=D=0$.
Am I correct or are they? And if my answer is incorrect how does one of the $x$'s cancel out from the $(Cx+D)$ term? Thanks for any help!
 [/INST] You are right. Here's an other proof:
If we put $X=x^2$ then
$$f(x)=\frac{1}{x^2(25+x^2)}$$
$$=\frac{1}{X(25+X)}$$
$$=\frac{B}{X}+\frac{D}{25+X}$$
$$Xf(x)=B+\frac{DX}{X+25}$$
with $ X=0$, it gives $ B=\frac{1}{25}$
$$(X+25)f(x)=\frac{B(X+25)}{X}+D$$
with $ X=-25$, we get $D=-\frac{1}{25}$
thus
$$f(x)=\frac{1}{25}\Bigl(\frac{1}{x^2}-\frac{1}{25+x^2}\Bigr)$$
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