Solve the integral of (x+3)/(x(x2+4)) using partial fraction decomposition.

First, factor the denominator $x^3+4x$:

$$x^3+4x = x(x^2+4)$$

Since $x^2+4$ is an irreducible quadratic factor, the partial fraction decomposition for the integrand is:

$$\frac{x+3}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4}$$

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

$$x+3 = A(x^2+4) + (Bx+C)x$$

Expand the right side:

$$x+3 = Ax^2+4A + Bx^2+Cx$$

Group terms by powers of $x$:

$$x+3 = (A+B)x^2 + Cx + 4A$$

Equate the coefficients of corresponding powers of $x$ on both sides:
For $x^2$: $A+B = 0 \quad (1)$
For $x$: $C = 1 \quad (2)$
For the constant term: $4A = 3 \quad (3)$

From equation (3), we find $A$:

$$A = \frac{3}{4}$$

From equation (2), we find $C$:

$$C = 1$$

Substitute the value of $A$ into equation (1):

$$\frac{3}{4} + B = 0 \implies B = -\frac{3}{4}$$

The constants are $A = \frac{3}{4}$, $B = -\frac{3}{4}$, and $C = 1$. These match the values given in the image.

Substitute the values of $A$, $B$, and $C$ back into the partial fraction decomposition:

$$\frac{x+3}{x(x^2+4)} = \frac{3/4}{x} + \frac{(-3/4)x+1}{x^2+4}$$

This can be split into three simpler fractions:

$$\frac{x+3}{x(x^2+4)} = \frac{3}{4x} - \frac{3x}{4(x^2+4)} + \frac{1}{x^2+4}$$

Now, we can integrate each term.

The integral becomes:

$$\int \left( \frac{3}{4x} - \frac{3x}{4(x^2+4)} + \frac{1}{x^2+4} \right) dx$$ $$= \int \frac{3}{4x} dx - \int \frac{3x}{4(x^2+4)} dx + \int \frac{1}{x^2+4} dx$$

Integrate the first term:

$$\int \frac{3}{4x} dx = \frac{3}{4} \int \frac{1}{x} dx = \frac{3}{4} \ln|x|$$

Integrate the second term. Let $u = x^2+4$, so $du = 2x \, dx \implies x \, dx = \frac{1}{2} du$:

$$\int \frac{3x}{4(x^2+4)} dx = \frac{3}{4} \int \frac{1}{x^2+4} (x \, dx) = \frac{3}{4} \int \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{3}{8} \int \frac{1}{u} du = \frac{3}{8} \ln|u| = \frac{3}{8} \ln(x^2+4)$$

Integrate the third term using the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right)$. Here $a^2=4$, so $a=2$:

$$\int \frac{1}{x^2+4} dx = \frac{1}{2} \arctan\left(\frac{x}{2}\right)$$

Combine the results from integrating each term and add the constant of integration $C$:

$$\int \frac{x+3}{x^3+4x} dx = \frac{3}{4} \ln|x| - \frac{3}{8} \ln(x^2+4) + \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C$$

The final answer is $\boxed{\frac{3}{4} \ln|x| - \frac{3}{8} \ln(x^2+4) + \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C}$.