Evaluate the integral: I = integral (x³)/((x²-1)²) dx

Evaluate the integral: $$ I = \int \frac{x^3}{(x^2-1)^2}\,dx. $$

Step 1 · Substitute

Let $u=x^2-1$, $du=2x\,dx$, so $x\,dx=\tfrac12du$ and $x^2=u+1$:

$$ x^3\,dx = x^2\cdot x\,dx = (u+1)\cdot\frac{du}{2}. $$

Step 2 · Rewrite the integral

$$ I = \int \frac{(u+1)/2}{u^2}\,du = \frac12\int \left(\frac1u+\frac{1}{u^2}\right) du. $$

Step 3 · Integrate

$$ I = \frac12\left(\ln|u| - \frac1u\right) + C. $$

Step 4 · Substitute back

With $u=x^2-1$:

$$ I = \frac12\ln\left|x^2-1\right| - \frac{1}{2(x^2-1)} + C. $$

Summary Table

Additional Notes

• The domain excludes $x=\pm1$, where the denominator vanishes.
• Writing $x^2=u+1$ to express the leftover $x^2$ factor from $x^3\,dx=x^2\cdot x\,dx$ entirely in terms of $u$ is the key step that avoids a full partial-fraction decomposition of $(u+1)/u^2$ into separate cover-up coefficients — it splits directly into $1/u+1/u^2$.
• The result can be checked by differentiating $\tfrac12\ln|x^2-1|-\tfrac{1}{2(x^2-1)}$: this gives $\tfrac{x}{x^2-1}+\tfrac{x}{(x^2-1)^2}=\tfrac{x(x^2-1)+x}{(x^2-1)^2}=\tfrac{x^3}{(x^2-1)^2}$, matching the integrand.