Evaluate the integral: I = integral x ln (x²-1) dx
Evaluate the integral: $$ I = \int x\ln\left(x^2-1\right) dx. $$
Explanation of Method
Integration by parts with $u=\ln(x^2-1)$, $dv=x\,dx$ leaves an improper rational function $x^3/(x^2-1)$, which is handled by polynomial division into a polynomial plus a standard logarithmic term.
- Choose $u=\ln(x^2-1)$, $dv=x\,dx$.
- Differentiate $u$ and integrate $dv$.
- Divide the leftover improper fraction.
- Integrate each piece and combine.
Step 1 · Set up integration by parts
Let:
$$ u = \ln(x^2-1) \quad \Rightarrow \quad du = \frac{2x}{x^2-1}\,dx, \qquad dv = x\,dx \quad \Rightarrow \quad v = \frac{x^2}{2}. $$
Step 2 · Apply the formula
$$ I = \frac{x^2}{2}\ln(x^2-1) - \int \frac{x^2}{2}\cdot\frac{2x}{x^2-1}\,dx = \frac{x^2}{2}\ln(x^2-1) - \int \frac{x^3}{x^2-1}\,dx. $$
Step 3 · Divide the improper fraction
Since $x^3 = x(x^2-1)+x$, polynomial division gives:
$$ \frac{x^3}{x^2-1} = x + \frac{x}{x^2-1}. $$
So the leftover integral splits as:
$$ \int \frac{x^3}{x^2-1}\,dx = \int x\,dx + \int \frac{x}{x^2-1}\,dx = \frac{x^2}{2} + \frac12\ln|x^2-1|. $$
Step 4 · Combine the results
$$ I = \frac{x^2}{2}\ln(x^2-1) - \frac{x^2}{2} - \frac12\ln|x^2-1| + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $uv$ term | $\tfrac{x^2}{2}\ln(x^2-1)$ |
| $x$ | $x^2/2$ |
| $x/(x^2-1)$ | $\tfrac12\ln|x^2-1|$ |
Additional Notes
- The domain requires $|x|>1$, where $x^2-1>0$ so the logarithm is defined.
- The leftover fraction $x^3/(x^2-1)$ is improper (numerator degree $\geq$ denominator degree), which is the signal to divide first, exactly as in Problem 114 — skipping this step and integrating $x^3/(x^2-1)$ directly would be far more error-prone.
- The result can be checked by differentiating $\tfrac{x^2}{2}\ln(x^2-1)-\tfrac{x^2}{2}-\tfrac12\ln|x^2-1|$ and confirming it returns $x\ln(x^2-1)$ (the $x^3/(x^2-1)$ and $-x-x/(x^2-1)$ pieces cancel exactly).
Result
The solution to the integral is: