Evaluate the integral: I = integral (x^6)/((x²-5)^4) dx

Evaluate the integral: $$ I = \int \frac{x^6}{\left(x^2-5\right)^4}\,dx. $$

Step 1 · Rewrite the numerator

Let $v=x^2-5$, so $x^2=v+5$ and $x^6=(v+5)^3=v^3+15v^2+75v+125$. Then:

$$ \frac{x^6}{(x^2-5)^4} = \frac1{x^2-5} + \frac{15}{(x^2-5)^2} + \frac{75}{(x^2-5)^3} + \frac{125}{(x^2-5)^4}. $$

Step 2 · Reduction formula

With $a^2=5$, the reduction formula $\displaystyle J_n=\int\frac{dx}{(x^2-a^2)^n}=-\frac{x}{2(n-1)a^2(x^2-a^2)^{n-1}} - \frac{2n-3}{2(n-1)a^2}J_{n-1}$ gives, starting from $J_1=\dfrac1{2\sqrt5}\ln\left|\dfrac{x-\sqrt5}{x+\sqrt5}\right|$:

$$ J_2 = -\frac{x}{10(x^2-5)} - \frac1{10}J_1, \qquad J_3 = -\frac{x}{20(x^2-5)^2} - \frac3{20}J_2, \qquad J_4 = -\frac{x}{30(x^2-5)^3} - \frac16J_3. $$

Step 3 · Assemble $I=J_1+15J_2+75J_3+125J_4$

Substituting $J_2,J_3,J_4$ in terms of $J_1$ and the rational pieces, then collecting the coefficient of $J_1$ and of each power of $x/(x^2-5)^k$:

$$ \text{coefficient of } J_1: \quad 1-\frac32+\frac98-\frac5{16} = \frac5{16}. $$

$$ \text{coefficient of } \frac{x}{x^2-5}: -\frac32+\frac98-\frac5{16}=-\frac{11}{16}, \qquad \frac{x}{(x^2-5)^2}: -\frac{15}4+\frac{25}{24}=-\frac{65}{24}, \qquad \frac{x}{(x^2-5)^3}: -\frac{25}6. $$

Step 4 · Combine

Since $\tfrac5{16}J_1=\tfrac5{16}\cdot\tfrac1{2\sqrt5}\ln|\cdots|=\tfrac{\sqrt5}{32}\ln|\cdots|$:

$$ I = \frac{\sqrt5}{32}\ln\left|\frac{x-\sqrt5}{x+\sqrt5}\right| - \frac{11x}{16(x^2-5)} - \frac{65x}{24(x^2-5)^2} - \frac{25x}{6(x^2-5)^3} + C. $$

Summary Table

Additional Notes

• Correction: the source's final boxed answer used coefficients $\tfrac1{2\sqrt5}$ for the logarithm and $-\tfrac{3x}{2(x^2-5)}$, $-\tfrac{15x}{4(x^2-5)^2}$ for the first two rational terms — these do not satisfy the reduction formula correctly (the source appears to have applied the reduction formula to each power of $J_n$ independently without properly substituting the lower-order results back in, so the $J_1$ contributions from $J_2$, $J_3$, and $J_4$ were never combined). Differentiating the source's proposed answer at $x=3$ gives $\approx4.251$, while the true integrand value there is $729/256\approx2.848$ — confirming the error. The version above rebuilds $J_2,J_3,J_4$ explicitly in terms of $J_1$ before combining, giving the corrected coefficients $\sqrt5/32,\ -11/16,\ -65/24,\ -25/6$.
• The domain excludes $x=\pm\sqrt5$, where the denominator vanishes.
• The result was verified by differentiating the boxed expression and confirming, after combining over the common denominator $48(x^2-5)^4$, that the numerator reduces exactly to $48x^6$ — and independently, by numerically checking the derivative at $x=3$, which gives $729/256$, matching the integrand.