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!

Since the central theme is to assist in the evaluation of an integral another method is to consider the following.
$$ \frac{1}{x^2 \, (x^2 + 5^2)} = \frac{1}{25} \, \left( \frac{1}{x^2} - \frac{1}{x^2 + 5^2} \right) $$
and
\begin{align}
\frac{1}{x^2 + 5^2} &= \frac{A}{x - 5 \, i} - \frac{B}{x + 5 \, i} \\
&= \frac{(A-B) \, x + (A + B) \, 5 \, i}{x^2 + 5^2} \\
&= \frac{1}{10 \, i} \, \left( \frac{1}{x - 5 \, i} - \frac{1}{x + 5 \, i} \right).
\end{align}
Now,
$$ \frac{1}{x^2 \, (x^2 + 5^2)} = \frac{1}{25 \, x^2} - \frac{1}{250 \, i} \, \left( \frac{1}{x - 5 \, i} - \frac{1}{x + 5 \, i} \right) $$
and
\begin{align}
\int \frac{dx}{x^2 \, (x^2 + 5^2)} &= \int \left( \frac{1}{25 \, x^2} - \frac{1}{250 \, i} \, \left( \frac{1}{x - 5 \, i} - \frac{1}{x + 5 \, i} \right) \right) \, dx \\
&= - \frac{1}{25 \, x} + \frac{i}{250} \, \ln\left(\frac{x - 5 \, i}{x + 5 \, i} \right) \\
&= \frac{1}{5^3} \, \left( \tan^{-1}\left(\frac{5}{x}\right) - \frac{5}{x} \right) + c_{0} \\
&= \frac{1}{5^3} \, \left( \cot^{-1}\left(\frac{x}{5}\right) - \frac{5}{x} \right) + c_{0}
\end{align}
