Answer
$$\int\frac{dx}{1-x^2}=-\frac{1}{2}\ln|1-x|+\frac{1}{2}\ln|1+x|+C$$
Work Step by Step
$$A=\int\frac{dx}{1-x^2}$$
1) Express the integrand as a sum of partial fractions:
$$\frac{1}{1-x^2}=\frac{1}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x}$$
Clear fractions:
$$1=A(1+x)+B(1-x)$$ $$1=A+Ax+B-Bx$$ $$1=(A+B)+(A-B)x$$
Equating coefficients of corresponding powers of $x$, we get
- $A+B=1$
- $A-B=0$
Calculation gives us $A=B=1/2$
Therefore, $$\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$$
2) Evaluate the integral: $$A=\frac{1}{2}\int\frac{dx}{1-x}+\frac{1}{2}\int\frac{dx}{1+x}$$ $$A=-\frac{1}{2}\int\frac{d(1-x)}{1-x}+\frac{1}{2}\int\frac{d(1+x)}{1+x}$$ $$A=-\frac{1}{2}\ln|1-x|+\frac{1}{2}\ln|1+x|+C$$
