91Ó°ÊÓ

First, we will factor out 8 from the numerator: $$\int \frac{8\left(x^{2}+4\right)}{x\left(x^{2}+8\right)} d x = 8 \int \frac{x^{2}+4}{x\left(x^{2}+8\right)} d x$$

We will use partial fraction decomposition to solve the integral. Since the denominator contains a linear factor (\(x\)) and a quadratic factor (\(x^2+8\)), we will write the decomposed fraction as: $$\frac{x^{2}+4}{x\left(x^{2}+8\right)} = \frac{A}{x} + \frac{Bx + C}{x^{2}+8}$$

To find the constants \(A\), \(B\), and \(C\), we will clear the fractions by multiplying both sides by the denominator: $$(x^{2}+4) = Ax(x^{2}+8) + (Bx+C)x$$ Now we will choose some values for \(x\) to solve for the constants. Setting \(x = 0\), we get: $$0^2 + 4 = 8A$$ So, \(A = \frac{1}{2}\). Now, differentiate both sides and set \(x=0\) again: $$2x = 2Ax^2 + (2Bx + C)$$ $$0 = 0 + C$$ So, \(C = 0\). Now we are left with finding the value of \(B\): $$(x^2 + 4) = \frac{1}{2}x(x^2 + 8) + Bx^2$$ Comparing the coefficients of \(x^2\) on both sides, we get: $$1 = \frac{1}{2} + B$$ So, \(B = \frac{1}{2}\). Now, the fraction decomposition becomes: $$\frac{x^{2}+4}{x\left(x^{2}+8\right)} = \frac{1/2}{x} + \frac{1/2 x}{x^{2}+8}$$

Now we will integrate term by term: $$8 \int \frac{x^{2}+4}{x\left(x^{2}+8\right)} d x = 8\left( \int \frac{1/2}{x} dx + \int \frac{1/2 x}{x^{2}+8} dx\right)$$ $$= 4 \int\frac{1}{x} dx + 4 \int\frac{x}{x^2+8} dx$$ The first integral is trivial: $$4\int \frac{1}{x}dx = 4\ln|x| + C_1$$ Now let's solve the second integral using substitution. Let \(u=x^2+8\), so \(du=2x\,dx\): $$4\int \frac{x}{x^2+8} dx = 2\int \frac{du}{u}$$ $$= 2\ln|u| + C_2 = 2\ln|x^2+8| + C_2$$

Combine both results and obtain the final answer: $$8 \int \frac{x^{2}+4}{x\left(x^{2}+8\right)} d x = 4\ln|x| + 2\ln|x^2+8| + C$$ Therefore, the integral is: $$\int \frac{8\left(x^{2}+4\right)}{x\left(x^{2}+8\right)} d x = 4\ln|x| + 2\ln|x^2+8| + C$$