Integrate the rational function: $\frac{2}{(1-x)(1+x^{2})}$

(N/A) Let $\frac{2}{(1-x)(1+x^{2})} = \frac{A}{1-x} + \frac{Bx+C}{1+x^{2}}$
Multiplying both sides by $(1-x)(1+x^{2})$,we get:
$2 = A(1+x^{2}) + (Bx+C)(1-x)$
$2 = A + Ax^{2} + Bx - Bx^{2} + C - Cx$
$2 = (A-B)x^{2} + (B-C)x + (A+C)$
Equating the coefficients of $x^{2}, x,$ and the constant term,we obtain:
$A-B = 0 \Rightarrow A = B$
$B-C = 0 \Rightarrow B = C$
$A+C = 2$
Substituting $A=B$ and $C=B$ into $A+C=2$,we get $B+B=2$,so $B=1$. Thus,$A=1$ and $C=1$.
Therefore,$\frac{2}{(1-x)(1+x^{2})} = \frac{1}{1-x} + \frac{x+1}{1+x^{2}}$
Integrating both sides:
$\int \frac{2}{(1-x)(1+x^{2})} dx = \int \frac{1}{1-x} dx + \int \frac{x}{1+x^{2}} dx + \int \frac{1}{1+x^{2}} dx$
$= -\int \frac{-1}{1-x} dx + \frac{1}{2} \int \frac{2x}{1+x^{2}} dx + \int \frac{1}{1+x^{2}} dx$
$= -\log|1-x| + \frac{1}{2} \log|1+x^{2}| + \tan^{-1}(x) + K$
Where $K$ is the constant of integration.