Find the indefinite integral of 2 divided by 1 minus x squared.

Step-by-step solution

1. 1

   Factor the denominator. $1-x^2=(1-x)(1+x)$: two distinct linear factors, which is exactly the case where partial fractions produces two simple constants. Note the order $(1-x)$, not $(x-1)$ — keeping the signs as written avoids an extra minus later.

2. 2

   Set up the partial-fraction split. Write

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

   and clear denominators:

   $2=A(1+x)+B(1-x)=(A+B)+(A-B)x$

3. 3

   Match coefficients. The left side has no $x$ term and constant $2$, so

   $A-B=0,\qquad A+B=2\quad\Rightarrow\quad A=B=1$

   Hence $\dfrac{2}{1-x^2}=\dfrac{1}{1-x}+\dfrac{1}{1+x}$.

4. 4

   Integrate each piece, watching the inner derivative.

   $\int\frac{dx}{1-x}=-\ln|1-x|,\qquad \int\frac{dx}{1+x}=\ln|1+x|$

   The minus sign on the first comes from the chain rule: the inside $1-x$ has derivative $-1$. Forgetting it is the main pitfall of this integral.

5. 5

   Combine the logarithms.

   $\int\frac{2}{1-x^2}\,dx=\ln|1+x|-\ln|1-x|+C=\ln\left|\frac{1+x}{1-x}\right|+C$

   On the interval $|x|<1$ this is the same as $2\operatorname{artanh}(x)+C$.

6. 6

   Differentiate to check.

   $\frac{d}{dx}\left(\ln|1+x|-\ln|1-x|\right)=\frac{1}{1+x}+\frac{1}{1-x}=\frac{(1-x)+(1+x)}{(1+x)(1-x)}=\frac{2}{1-x^2}$

   The original integrand is recovered, confirming the answer.