We are asked to evaluate the integral:

$I = \int \frac{2x}{(x^2 + 1)(x^2 + 3)} \, dx$

Step 1: Using substitution to simplify

Let us use the substitution $u = x^2$. Then:

$du = 2x \, dx$

So, the integral becomes:

$I = \int \frac{du}{(u + 1)(u + 3)}$

Step 2: Decomposing into partial fractions

We now decompose the integrand into partial fractions:

$\frac{1}{(u + 1)(u + 3)} = \frac{A}{u + 1} + \frac{B}{u + 3}$

Multiplying both sides by $(u + 1)(u + 3)$, we get:

$1 = A(u + 3) + B(u + 1)$

Expanding this expression:

$1 = A(u) + 3A + B(u) + B$

$1 = (A + B)u + (3A + B)$

Step 3: Solving for $A$ and $B$

For the equation $1 = (A + B)u + (3A + B)$ to hold for all $u$, we equate the coefficients of $u$ and the constant terms:

1. $A + B = 0$
2. $3A + B = 1$

From the first equation, we have $B = -A$. Substituting into the second equation:

$3A - A = 1$

$2A = 1 \implies A = \frac{1}{2}$

Thus, $B = -\frac{1}{2}$.

Step 4: Substituting back into the integral

We now substitute these values of $A$ and $B$ into the partial fractions:

$\frac{1}{(u + 1)(u + 3)} = \frac{1/2}{u + 1} - \frac{1/2}{u + 3}$

So the integral becomes:

$I = \frac{1}{2} \int \left( \frac{1}{u + 1} - \frac{1}{u + 3} \right) du$

Step 5: Integrating

Now we integrate each term:

$I = \frac{1}{2} \left( \ln|u + 1| - \ln|u + 3| \right) + C$

$I = \frac{1}{2} \ln \left( \frac{|u + 1|}{|u + 3|} \right) + C$

Step 6: Substituting back $u = x^2$

Substituting $u = x^2$ back into the expression:

$I = \frac{1}{2} \ln \left( \frac{x^2 + 1}{x^2 + 3} \right) + C$

Thus, the integral is:

$\int \frac{2x}{(x^2 + 1)(x^2 + 3)} \, dx = \frac{1}{2} \ln \left( \frac{x^2 + 1}{x^2 + 3} \right) + C$

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Would you like more details on any step? Here are some related questions:

1. How do you perform partial fraction decomposition?
2. What is the method of substitution in integration?
3. Can you explain the properties of logarithms used in integrals?
4. What are common integrals involving rational functions?
5. How can you integrate more complex rational functions?

Tip: Always check if substitution simplifies a complex expression before starting with advanced techniques like partial fractions!