Evaluate the integral: I = integral 1/(1-x²) ln
Evaluate the integral: $$ I = \int \frac{1}{1-x^2} \ln \frac{1+x}{1-x} \, dx. $$
Explanation of Method
The key observation is that $\dfrac{d}{dx}\ln\dfrac{1+x}{1-x} = \dfrac{1}{1+x}+\dfrac{1}{1-x} = \dfrac{2}{1-x^2}$, which is (up to a constant factor of $2$) exactly the other factor multiplying the logarithm. Substituting $t=\ln\dfrac{1+x}{1-x}$ directly collapses the whole integral into $\int t\,dt$.
- Substitute $t=\ln\dfrac{1+x}{1-x}$ directly.
- Compute $dt$ and recognize it matches $\dfrac{2\,dx}{1-x^2}$.
- Rewrite the integral as $\int t\,dt$.
- Integrate and substitute back.
Step 1 · Substitute the logarithm itself
Rather than substituting the ratio $\dfrac{1+x}{1-x}$, let $t$ be the logarithm itself:
$$ t = \ln\frac{1+x}{1-x} = \ln(1+x) - \ln(1-x). $$
Differentiating:
$$ \frac{dt}{dx} = \frac{1}{1+x} + \frac{1}{1-x} = \frac{(1-x)+(1+x)}{(1+x)(1-x)} = \frac{2}{1-x^2}. $$
So:
$$ dt = \frac{2\,dx}{1-x^2} \quad \Rightarrow \quad \frac{dx}{1-x^2} = \frac{dt}{2}. $$
Step 2 · Rewrite the integral
The original integral is exactly $t$ times $\dfrac{dx}{1-x^2}$:
$$ I = \int \ln\frac{1+x}{1-x} \cdot \frac{dx}{1-x^2} = \int t\cdot \frac{dt}{2} = \frac{1}{2}\int t\,dt. $$
Step 3 · Integrate
Using the power rule:
$$ \frac{1}{2}\int t\,dt = \frac{1}{2}\cdot \frac{t^2}{2} = \frac{t^2}{4}. $$
Step 4 · Substitute back
Substituting back $t = \ln\dfrac{1+x}{1-x}$, we obtain:
$$ I = \frac{1}{4}\left(\ln\frac{1+x}{1-x}\right)^2 + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $t = \ln\dfrac{1+x}{1-x}$ |
| $dx/(1-x^2)$ in terms of $dt$ | $\tfrac12 dt$ |
| Antiderivative | $t^2/4$ |
Additional Notes
- The domain of $I$ requires $-1 < x < 1$, so that $\dfrac{1+x}{1-x}>0$ and $1-x^2 \neq 0$.
- Substituting the entire logarithmic expression $t=\ln\dfrac{1+x}{1-x}$ directly (rather than first setting $t$ equal to the ratio inside) is the key simplification — it avoids introducing extra algebra to convert between $t$, $x$, and the logarithm at multiple stages.
- The result can be checked by differentiating $\tfrac14\left(\ln\dfrac{1+x}{1-x}\right)^2$ with the chain rule and confirming it returns $\dfrac{1}{1-x^2}\ln\dfrac{1+x}{1-x}$.
Result
The solution to the integral is: