Question
Integrate the rational function $\frac{1}{x\left(x^{4}-1\right)}$

Answer

Given the function, $\frac{1}{x\left(x^{4}+1\right)}$ 
Multiplying numerator and denominator by x3, we get,
$\frac{1}{x\left(x^{4}+1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}+1\right)}$ 
Therefore, $\int \frac{1}{x\left(x^{4}+1\right)} d x=\int \frac{x^{3}}{x^{4}\left(x^{4}+1\right)} d x$ 
Now, let x4 = t
4x3dx = dt
Thus, $\int \frac{1}{x\left(x^{4}+1\right)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)}$ 
Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$ 
1 = A(t - 1) + Bt …(i)
Substituting t = 0 and 1 in (i), we get
A = -1 and B = 1
Therefore $\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$ 
$\int \frac{1}{x\left(x^{4}+1\right)} d x=\frac{1}{4} \int \left\{\frac{-1}{t}+\frac{1}{t-1}\right\} d t$ 
= $\frac{1}{4}[-\log |t|+\log |t-1|]+C$ 
= $\frac{1}{4} \log \left|\frac{t-1}{t}\right|+C$ 
= $\frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+C$

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