Evaluate the integral: I = integral (x²)/(1-x^4) dx

Evaluate the integral: $$ I = \int \frac{x^2}{1-x^4}\,dx. $$

Step 1 · Factor and set up partial fractions

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

Multiplying through: $x^2=A(1+x^2)+B(1-x^2)$. Matching coefficients gives $A-B=1$ and $A+B=0$, so $A=\tfrac12$, $B=-\tfrac12$.

Step 2 · Integrate each piece

Using $1-x^2=(1-x)(1+x)$, the standard result is $\int dx/(1-x^2)=\tfrac12\ln\left|\tfrac{1+x}{1-x}\right|$, and $\int dx/(1+x^2)=\arctan x$:

$$ I = \frac12\cdot\frac12\ln\left|\frac{1+x}{1-x}\right| - \frac12\arctan x + C. $$

Step 3 · Simplify

$$ I = \frac14\ln\left|\frac{1+x}{1-x}\right| - \frac12\arctan x + C. $$

Summary Table

Additional Notes

• The domain excludes $x=\pm1$, where $1-x^2$ vanishes.
• This mirrors Problem 84's family: splitting an even quartic-denominator rational function into a "difference of squares" pair, one bounded ($1-x^2$, giving a log) and one unbounded ($1+x^2$, giving an arctangent).
• The result can be checked by differentiating $\tfrac14\ln\left|\tfrac{1+x}{1-x}\right|-\tfrac12\arctan x$: the first term gives $\tfrac{1}{2(1-x^2)}$ (using $\tfrac{d}{dx}\ln\left|\tfrac{1+x}{1-x}\right|=\tfrac{1}{1+x}+\tfrac{1}{1-x}=\tfrac{2}{1-x^2}$), and the second gives $-\tfrac{1}{2(1+x^2)}$; combining over $2(1-x^4)$ gives numerator $(1+x^2)-(1-x^2)=2x^2$, reducing to $\tfrac{2x^2}{2(1-x^4)}=\tfrac{x^2}{1-x^4}$, matching the integrand.