Evaluate the integral: I = integral x²ln ((1-x)/(1+x)) dx
Evaluate the integral: $$ I = \int x^2\ln\left(\frac{1-x}{1+x}\right) dx. $$
Explanation of Method
Integration by parts with $u=\ln\!\left(\tfrac{1-x}{1+x}\right)$, $dv=x^2\,dx$ produces a leftover rational integral $\int\tfrac{x^3}{1-x^2}\,dx$, which is resolved by the substitution $t=1-x^2$.
- Choose $u=\ln\!\left(\tfrac{1-x}{1+x}\right)$, $dv=x^2\,dx$.
- Differentiate $u$ and integrate $dv$.
- Evaluate the leftover integral via $t=1-x^2$.
- Combine the results, keeping the sign from integration by parts.
Step 1 · Set up integration by parts
Let:
$$ u = \ln\left(\frac{1-x}{1+x}\right) = \ln(1-x)-\ln(1+x), \qquad dv = x^2\,dx \quad \Rightarrow \quad v = \frac{x^3}{3}. $$
Differentiating $u$:
$$ du = \left(\frac{-1}{1-x} - \frac{1}{1+x}\right) dx = \frac{-2}{1-x^2}\,dx. $$
Step 2 · Apply the formula
$$ I = \frac{x^3}{3}\ln\left(\frac{1-x}{1+x}\right) - \int \frac{x^3}{3}\cdot\frac{-2}{1-x^2}\,dx = \frac{x^3}{3}\ln\left(\frac{1-x}{1+x}\right) + \frac23\int \frac{x^3}{1-x^2}\,dx. $$
Step 3 · Evaluate the leftover integral
Let $t=1-x^2$, so $dt=-2x\,dx$ and $x^2=1-t$:
$$ \int \frac{x^3}{1-x^2}\,dx = \int \frac{x^2\cdot x\,dx}{t} = \int \frac{(1-t)}{t}\cdot\left(-\frac{dt}{2}\right) = -\frac12\int \left(\frac1t-1\right) dt = -\frac12\left(\ln|t|-t\right). $$
Substituting back $t=1-x^2$:
$$ \int \frac{x^3}{1-x^2}\,dx = -\frac12\ln|1-x^2| + \frac{1-x^2}{2}. $$
Step 4 · Combine the results
Substituting into the expression for $I$:
$$ I = \frac{x^3}{3}\ln\left(\frac{1-x}{1+x}\right) + \frac23\left[-\frac12\ln|1-x^2| + \frac{1-x^2}{2}\right] + C. $$
Simplifying:
$$ I = \frac{x^3}{3}\ln\left(\frac{1-x}{1+x}\right) - \frac13\ln|1-x^2| + \frac{1-x^2}{3} + C. $$
Summary Table
| Piece | Value |
|---|---|
| $uv$ term | $\tfrac{x^3}{3}\ln\!\left(\tfrac{1-x}{1+x}\right)$ |
| $\int x^3/(1-x^2)\,dx$ | $-\tfrac12\ln|1-x^2|+\tfrac{1-x^2}{2}$ |
Additional Notes
- The domain requires $-1<x<1$, where $(1-x)/(1+x)>0$ so the logarithm is defined.
- A common pitfall in this problem is losing track of the overall sign when substituting the leftover integral back into $I = uv-\int v\,du$: since $du$ already carries a minus sign, the leftover integral enters with a plus sign, and it's easy to accidentally flip it back to minus when writing the final combined answer.
- The result can be checked by differentiating the final expression and confirming the polynomial and rational pieces cancel exactly, leaving only $x^2\ln\!\left(\tfrac{1-x}{1+x}\right)$.
Result
The solution to the integral is: