Evaluate the integral: I = integral ((x+1)³)/(x²-1) dx

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

Step 1 · Cancel the common factor

Since $x^2-1=(x-1)(x+1)$:

$$ \frac{(x+1)^3}{x^2-1} = \frac{(x+1)^3}{(x-1)(x+1)} = \frac{(x+1)^2}{x-1}. $$

Step 2 · Polynomial long division

$$ \frac{(x+1)^2}{x-1} = \frac{x^2+2x+1}{x-1} = x+3+\frac{4}{x-1}. $$

Step 3 · Integrate

$$ I = \int (x+3)\,dx + \int \frac{4}{x-1}\,dx = \frac{x^2}{2}+3x + 4\ln|x-1| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=1$ (note $x=-1$ is a removable singularity of the original expression, since the $(x+1)$ factor cancels — the simplified integrand $(x+1)^2/(x-1)$ is perfectly well-defined there).
• Spotting the common factor $(x+1)$ between numerator and denominator before expanding avoids unnecessary polynomial expansion of $(x+1)^3$ and a longer division — a useful shortcut whenever the numerator shares a visible factor with the denominator.
• The result can be checked by differentiating $\tfrac{x^2}2+3x+4\ln|x-1|$: this gives $x+3+\tfrac{4}{x-1}=\tfrac{(x+3)(x-1)+4}{x-1}=\tfrac{x^2+2x+1}{x-1}=\tfrac{(x+1)^2}{x-1}$, which equals the original integrand $(x+1)^3/(x^2-1)$ after multiplying numerator and denominator by $(x+1)$.