<s>[INST] $\int 1/(x^3-25x) \mathrm{d}x$ Integrate $\displaystyle \frac{1}{x^3-25x}$
Between limits of $4,3$.
Think it involves the 'Difference in squares' substitution.
I ended up with an answer of $0.3242$ which is wrong as it is $0.023$.
Any help and pointers would be greatly appreciated.
Matt
 [/INST] John's approach is correct. 
$\frac{1}{x^3-25x} = \frac{1}{x(x+5)(x-5)} = \frac{A}{x} + \frac{B}{x+5} + \frac{C}{x-5}$.
Continuing, we generate the following equation: $$A(x+5)(x-5)+Bx(x-5)+Cx(x+5)=1$$
By strategically choosing our $x$ values (I recommend $x=-5,0,$ and $5$), we can derive that $A=\frac {-1}{25}, B= \frac{1}{50},$ and $C= \frac {1}{50}$. Thus, we rewrite:
$\int \frac {1}{x^3-25x}= \int \frac{-1/25}{x} + \frac{1/50}{x+5} + \frac{1/50}{x-5} \mathrm{d} x$.
I suggest breaking the integral into three separate integrals and proceed by setting the denominator of each equal to some dummy variable. Then, all three very nicely fit the form $\int \frac {\mathrm{d} u} {u}$. Can you proceed from here?
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