Evaluate the integral: I = integral (x^4-1)/(x (x^4-5)

Evaluate the integral: $$ I = \int \frac{x^4-1}{x\left(x^4-5\right)\left(x^5-5x+1\right)}\,dx. $$

Step 1 · Rewrite the denominator

Let $w=x^5-5x=x(x^4-5)$. Then:

$$ x^5-5x+1 = w+1, \qquad x(x^4-5)(x^5-5x+1) = w(w+1). $$

Step 2 · Substitute

Since $\dfrac{dw}{dx}=5x^4-5=5(x^4-1)$, we have $(x^4-1)\,dx=\dfrac{dw}5$. The integral becomes:

$$ I = \int \frac{(x^4-1)\,dx}{w(w+1)} = \frac15\int\frac{dw}{w(w+1)}. $$

Step 3 · Partial fractions in $w$

$$ \frac1{w(w+1)} = \frac1w-\frac1{w+1} \quad\Rightarrow\quad \frac15\int\left(\frac1w-\frac1{w+1}\right)dw = \frac15\ln\left|\frac{w}{w+1}\right| + C. $$

Step 4 · Back-substitute

$$ I = \frac15\ln\left|\frac{x(x^4-5)}{x^5-5x+1}\right| + C. $$

Summary Table

Additional Notes

• Major rewrite: the source attempted a direct triple-log ansatz — $-\tfrac15\ln|x|+\tfrac1{20}\ln|x^4-5|+\tfrac15\ln|x^5-5x+1|$ — asserted without a real derivation ("given the complexity... we recognize the pattern"). Differentiating that expression at $x=2$ gives $\approx0.698$, while the true integrand value is $15/506\approx0.0296$: off by more than a factor of 20, so the source's answer is simply wrong. The key realization missed by the source is that $x(x^4-5)=x^5-5x$ is exactly the non-constant part of $x^5-5x+1$, which turns the whole three-factor denominator into $w(w+1)$ under the substitution $w=x^5-5x$ — a textbook partial-fraction integral once that substitution is made.
• The domain excludes $x=0$, the real roots of $x^4=5$ (i.e. $x=\pm\sqrt[4]5$), and any real root of $x^5-5x+1=0$.
• The result was verified by differentiating $\tfrac15\ln|x(x^4-5)|-\tfrac15\ln|x^5-5x+1|$: this gives $\tfrac15\left[\tfrac1x+\tfrac{4x^3}{x^4-5}\right]-\tfrac{x^4-1}{x^5-5x+1}$, which evaluates to $0.681818-0.652174=0.029644$ at $x=2$, matching the integrand value $15/506\approx0.029644$ exactly.