The question in the image asks to find the result of the integral:

$\int_0^2 \frac{6x}{x^2 + 6x + 8} \, dx$

Let's solve this step-by-step:

Step 1: Factor the denominator

First, we factor the quadratic expression in the denominator:

$x^2 + 6x + 8 = (x + 2)(x + 4)$

So the integral becomes:

$\int_0^2 \frac{6x}{(x + 2)(x + 4)} \, dx$

Step 2: Use Partial Fraction Decomposition

We need to express the integrand in terms of partial fractions. Let:

$\frac{6x}{(x + 2)(x + 4)} = \frac{A}{x + 2} + \frac{B}{x + 4}$

Multiplying both sides by $(x + 2)(x + 4)$ to clear the denominators:

$6x = A(x + 4) + B(x + 2)$

Now, expand the right-hand side:

$6x = A(x) + 4A + B(x) + 2B$ $6x = (A + B)x + (4A + 2B)$

Step 3: Set up the system of equations

Equating the coefficients of like terms:

1. For the $x$-terms: $A + B = 6$
2. For the constant terms: $4A + 2B = 0$

Step 4: Solve for $A$ and $B$

From the second equation, solve for $A$:

$4A + 2B = 0 \quad \Rightarrow \quad 2A + B = 0 \quad \Rightarrow \quad B = -2A$

Substitute $B = -2A$ into the first equation:

$A - 2A = 6 \quad \Rightarrow \quad -A = 6 \quad \Rightarrow \quad A = -6$

Then substitute $A = -6$ into $B = -2A$:

$B = -2(-6) = 12$

Step 5: Rewrite the integral

Now that we have $A = -6$ and $B = 12$, we can rewrite the integrand as:

$\int_0^2 \frac{-6}{x + 2} + \frac{12}{x + 4} \, dx$

Step 6: Integrate term by term

The integral of $\frac{1}{x + a}$ is $\ln|x + a|$. Therefore, we integrate each term:

$\int_0^2 \frac{-6}{x + 2} \, dx = -6 \ln|x + 2| \Big|_0^2 = -6 (\ln(4) - \ln(2)) = -6 \ln(2)$ $\int_0^2 \frac{12}{x + 4} \, dx = 12 \ln|x + 4| \Big|_0^2 = 12 (\ln(6) - \ln(4)) = 12 \ln\left(\frac{3}{2}\right)$

Step 7: Combine the results

Now combine the two parts:

$-6 \ln(2) + 12 \ln\left(\frac{3}{2}\right)$

This simplifies to:

$-6 \ln(2) + 12 (\ln(3) - \ln(2)) = -6 \ln(2) + 12 \ln(3) - 12 \ln(2) = -18 \ln(2) + 12 \ln(3)$

Step 8: Evaluate the expression

Plugging in approximate values for the logarithms:

$\ln(2) \approx 0.6931, \quad \ln(3) \approx 1.0986$

We get:

$-18(0.6931) + 12(1.0986) = -12.4758 + 13.1832 = 0.7074$

Thus, the value of the integral is approximately $0.707$.

Let me know if you'd like a more detailed explanation or have any further questions!