this problem is about integral involving $\frac{1}{u}du$\n$int\frac{6x^{2}+8}{4x + x^{3}}dx$

Explanation:

Step1: Let (u = 4x + x^{3})

Differentiate (u) with respect to (x). Using the power - rule ((x^n)^\prime=nx^{n - 1}), we have (du=(4 + 3x^{2})dx). Notice that (6x^{2}+8 = 2(3x^{2}+4)).

Step2: Rewrite the integral

(\int\frac{6x^{2}+8}{4x + x^{3}}dx=\int\frac{2(3x^{2}+4)}{4x + x^{3}}dx). Since (du=(3x^{2}+4)dx), the integral becomes (2\int\frac{du}{u}).

Step3: Integrate (\frac{1}{u})

We know that (\int\frac{1}{u}du=\ln|u|+C). So (2\int\frac{du}{u}=2\ln|u|+C).

Step4: Substitute back (u = 4x + x^{3})

The result is (2\ln|4x + x^{3}|+C).

Answer:

(2\ln|4x + x^{3}|+C)