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We will decompose the integrand into partial fractions. Let $$ \frac{2x}{(x^2+3)(x^2-5)} = \frac{A}{x^2+3} + \frac{B}{x^2-5} $$ Multiplying both sides by $(x^2+3)(x^2-5)$, we get $$ 2x = A(x^2-5) + B(x^2+3) $$ $$ 2x = (A+B)x^2 + (-5A+3B) $$ Comparing coefficients, we have $$ A+B = 0 \quad \text{and} \quad -5A+3B = 2x $$ Since the left side is $2x$, there seems to be an error. The correct equation should be $-5A+3B = 0$. However, the original equation is correct. Let's proceed with the correct equation $2x = A(x^2-5) + B(x^2+3)$. Since there is no $x^2$ term on the left side, $A+B = 0$, so $B = -A$. Since there is no constant term on the left side, $-5A+3B = 0$. Substituting $B = -A$, we get $-5A - 3A = 0$, so $-8A = 0$, which means $A = 0$ and $B = 0$. This is incorrect. The correct equation is $2x = A(x^2-5) + B(x^2+3)$. Let $x=0$. Then $0 = -5A + 3B$. Also, the coefficient of $x^2$ is $A+B = 0$, so $A = -B$. Then $0 = -5A - 3A = -8A$, so $A = 0$ and $B = 0$. This is still incorrect. Let's rewrite the equation as $$ \frac{2x}{(x^2+3)(x^2-5)} = \frac{Ax+C}{x^2+3} + \frac{Bx+D}{x^2-5} $$ Then $2x = (Ax+C)(x^2-5) + (Bx+D)(x^2+3)$ $2x = Ax^3 - 5Ax + Cx^2 - 5C + Bx^3 + 3Bx + Dx^2 + 3D$ $2x = (A+B)x^3 + (C+D)x^2 + (-5A+3B)x + (-5C+3D)$ Comparing coefficients, we have $A+B = 0$ $C+D = 0$ $-5A+3B = 2$ $-5C+3D = 0$ From $A+B = 0$, $A = -B$. Substituting into $-5A+3B = 2$, we get $5B+3B = 2$, so $8B = 2$, and $B = \frac{1}{4}$. Then $A = -\frac{1}{4}$. From $C+D = 0$, $C = -D$. Substituting into $-5C+3D = 0$, we get $5D+3D = 0$, so $8D = 0$, and $D = 0$. Then $C = 0$. Thus, $\frac{2x}{(x^2+3)(x^2-5)} = \frac{-\frac{1}{4}x}{x^2+3} + \frac{\frac{1}{4}x}{x^2-5}$
Now we integrate: $$ \int \frac{2x}{(x^2+3)(x^2-5)} dx = \int \left( \frac{-\frac{1}{4}x}{x^2+3} + \frac{\frac{1}{4}x}{x^2-5} \right) dx $$ $$ = -\frac{1}{4} \int \frac{x}{x^2+3} dx + \frac{1}{4} \int \frac{x}{x^2-5} dx $$ Let $u = x^2+3$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$. Let $v = x^2-5$, then $dv = 2x dx$, so $x dx = \frac{1}{2} dv$. $$ = -\frac{1}{4} \int \frac{1}{u} \frac{1}{2} du + \frac{1}{4} \int \frac{1}{v} \frac{1}{2} dv $$ $$ = -\frac{1}{8} \int \frac{1}{u} du + \frac{1}{8} \int \frac{1}{v} dv $$ $$ = -\frac{1}{8} \ln|u| + \frac{1}{8} \ln|v| + C $$ $$ = -\frac{1}{8} \ln|x^2+3| + \frac{1}{8} \ln|x^2-5| + C $$ $$ = \frac{1}{8} \ln \left| \frac{x^2-5}{x^2+3} \right| + C $$
Final Answer: $\frac{1}{8} \ln \left| \frac{x^2-5}{x^2+3} \right| + C$
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