Evaluate the integral: I = integral (x²)/(1 - x²) dx
Evaluate the integral: $$ I = \int \frac{x^2}{1 - x^2} \, dx. $$
Explanation of Method
When the degree of the numerator matches the degree of the denominator, the fraction is improper and should first be rewritten as a constant plus a proper fraction. The remaining proper fraction, $\frac{1}{1-x^2}$, is a standard integral evaluated using the difference-of-squares logarithmic formula.
- Rewrite the improper fraction as a constant plus a proper fraction.
- Split the integral into two simpler integrals.
- Evaluate each piece, using the standard form for $\int \frac{dx}{1-x^2}$.
- Combine the results and add the constant of integration.
Step 1 · Simplify the integrand
We rewrite the fraction $\dfrac{x^2}{1-x^2}$ by splitting it into simpler terms:
$$ \frac{x^2}{1-x^2} = -1 + \frac{1}{1-x^2}. $$
Thus, the integral becomes:
$$ I = \int \frac{x^2}{1-x^2}\,dx = \int \left(-1 + \frac{1}{1-x^2}\right) dx. $$
Step 2 · Separate the integral
We now split the integral into two parts:
$$ I = \int -1\,dx + \int \frac{1}{1-x^2}\,dx. $$
First term: the integral of $-1$ is:
$$ \int -1\,dx = -x. $$
Second term: the integral of $\dfrac{1}{1-x^2}$ is a standard integral:
$$ \int \frac{1}{1-x^2}\,dx = \frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|. $$
Step 3 · Combine the results
Combining the results of both terms, we have:
$$ I = -x + \frac{1}{2}\ln\left|\frac{1+x}{1-x}\right| + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Type | Antiderivative |
|---|---|---|
| $-1$ | Constant term | $-x$ |
| $1/(1-x^2)$ | Standard logarithmic form | $\dfrac{1}{2}\ln\left|\dfrac{1+x}{1-x}\right|$ |
Additional Notes
- The domain of $I$ requires $x \neq \pm 1$, since the denominator $1-x^2$ vanishes there.
- Whenever the numerator and denominator of a rational integrand share the same degree, polynomial division first (here, $x^2 = -(1-x^2)+1$) simplifies the integral considerably.
- The result can be checked by differentiating $-x + \dfrac{1}{2}\ln\left|\dfrac{1+x}{1-x}\right|$ and confirming it returns the original integrand.
Result
The solution to the integral is: