We need to compute the definite integral:

$\int_{1}^{3} \frac{2x + 1}{x^3 + 8x} \, dx.$

Step 1: Simplify the integrand

The denominator $x^3 + 8x$ can be factored as:

$x^3 + 8x = x(x^2 + 8).$

Thus, the integral becomes:

$\int_{1}^{3} \frac{2x + 1}{x(x^2 + 8)} \, dx.$

Step 2: Decompose into partial fractions

We write the fraction:

$\frac{2x + 1}{x(x^2 + 8)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 8}.$

Multiply through by the denominator $x(x^2 + 8)$ to get:

$2x + 1 = A(x^2 + 8) + (Bx + C)x.$

Expanding and grouping terms:

$2x + 1 = A x^2 + 8A + Bx^2 + Cx.$

Combine like terms:

$2x + 1 = (A + B)x^2 + Cx + 8A.$

Equating coefficients of $x^2$, $x$, and the constant terms:

1. $A + B = 0$,
2. $C = 2$,
3. $8A = 1 \implies A = \frac{1}{8}$.

From $A + B = 0$, we get $B = -\frac{1}{8}$.

Thus, the partial fraction decomposition is:

$\frac{2x + 1}{x(x^2 + 8)} = \frac{1/8}{x} + \frac{-\frac{1}{8}x + 2}{x^2 + 8}.$

Step 3: Rewrite the integral

Substitute the partial fractions:

$\int_{1}^{3} \frac{2x + 1}{x(x^2 + 8)} \, dx = \int_{1}^{3} \frac{1/8}{x} \, dx + \int_{1}^{3} \frac{-\frac{1}{8}x}{x^2 + 8} \, dx + \int_{1}^{3} \frac{2}{x^2 + 8} \, dx.$

Step 4: Solve each term

First term:

$\int_{1}^{3} \frac{1/8}{x} \, dx = \frac{1}{8} \int_{1}^{3} \frac{1}{x} \, dx = \frac{1}{8} \ln|x| \Big|_{1}^{3} = \frac{1}{8} (\ln 3 - \ln 1) = \frac{1}{8} \ln 3.$

Second term:

$\int_{1}^{3} \frac{-\frac{1}{8}x}{x^2 + 8} \, dx.$

Let $u = x^2 + 8$, so $du = 2x \, dx$. The limits for $u$ are:

• When $x = 1$, $u = 1^2 + 8 = 9$,
• When $x = 3$, $u = 3^2 + 8 = 17$.

The integral becomes:

$-\frac{1}{8} \int_{9}^{17} \frac{1}{u} \, du = -\frac{1}{8} \ln|u| \Big|_{9}^{17} = -\frac{1}{8} (\ln 17 - \ln 9) = -\frac{1}{8} \ln\left(\frac{17}{9}\right).$

Third term:

$\int_{1}^{3} \frac{2}{x^2 + 8} \, dx.$

Factor $x^2 + 8 = 2^2 + x^2$. The integral becomes:

$\int_{1}^{3} \frac{2}{x^2 + 2^2} \, dx = \int_{1}^{3} \frac{1}{x^2 + 2^2} \, dx.$

This is a standard arctangent formula:

$\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right).$

Here, $a = 2$, so:

$\int_{1}^{3} \frac{2}{x^2 + 8} \, dx = \frac{1}{2} \arctan\left(\frac{x}{2}\right) \Big|_{1}^{3}.$

Evaluate:

$\frac{1}{2} \left[\arctan\left(\frac{3}{2}\right) - \arctan\left(\frac{1}{2}\right)\right].$

Step 5: Combine results

The total integral is:

$\frac{1}{8} \ln 3 - \frac{1}{8} \ln\left(\frac{17}{9}\right) + \frac{1}{2} \left[\arctan\left(\frac{3}{2}\right) - \arctan\left(\frac{1}{2}\right)\right].$

Simplify:

$\frac{1}{8} \ln\left(\frac{3 \cdot 9}{17}\right) + \frac{1}{2} \left[\arctan\left(\frac{3}{2}\right) - \arctan\left(\frac{1}{2}\right)\right].$

This is the evaluated definite integral.

Let me know if you need further details or approximations! 😊

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Questions:

1. Do you need a numerical approximation for this integral?
2. Would you like a step-by-step breakdown for partial fraction decomposition?
3. Should I explain how arctangent integrals work in more detail?
4. Do you need help with substitution techniques in integration?
5. Would you like to verify the solution using numerical integration?

Tip:

When dealing with partial fractions, always double-check the decomposition coefficients to avoid errors in later steps!