Integrate the rational function: $\frac{2x}{(x^2+1)(x^2+3)}$

(A) Let $I = \int \frac{2x}{(x^2+1)(x^2+3)} dx$.
Substitute $x^2 = t$,then $2x dx = dt$.
Thus,the integral becomes $I = \int \frac{dt}{(t+1)(t+3)}$.
Using partial fractions,let $\frac{1}{(t+1)(t+3)} = \frac{A}{t+1} + \frac{B}{t+3}$.
This implies $1 = A(t+3) + B(t+1)$.
Setting $t = -1$,we get $1 = A(2) \Rightarrow A = \frac{1}{2}$.
Setting $t = -3$,we get $1 = B(-2) \Rightarrow B = -\frac{1}{2}$.
Therefore,$I = \int \left( \frac{1}{2(t+1)} - \frac{1}{2(t+3)} \right) dt$.
Integrating,we get $I = \frac{1}{2} \ln|t+1| - \frac{1}{2} \ln|t+3| + C$.
Substituting $t = x^2$ back,we get $I = \frac{1}{2} \ln \left| \frac{x^2+1}{x^2+3} \right| + C$,where $C$ is the constant of integration.