Partial Fraction Decomposition of $frac{1}{x^2(x^2+25)}$

Question

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: $intfrac{dx}{x^2(x^2+25)}$. 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!

Leucippus · Answer

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} , lnleft(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}

hamam_Abdallah · Answer

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
$$f(x)=frac{1}{25}Bigl(frac{1}{x^2}-frac{1}{25+x^2}Bigr)$$

Partial Fraction Decomposition of $frac{1}{x^2(x^2+25)}$

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