Certainly! Let's solve the integral
[tex]\[
\int \frac{\sqrt[3]{4 + 3 \ln (2x)}}{x} \, dx
\][/tex]
step-by-step.
### Step 1: Simplify the Integrand
We start with the given function:
[tex]\[
\frac{\sqrt[3]{4 + 3 \ln (2x)}}{x}
\][/tex]
### Step 2: Substitute
Let's use the substitution [tex]\( u = 4 + 3 \ln(2x) \)[/tex]. First, find [tex]\( du \)[/tex].
Taking the derivative [tex]\( du \)[/tex]:
[tex]\[
du = 3 \cdot \frac{1}{2x} \cdot 2dx = \frac{3}{x}dx
\][/tex]
Thus,
[tex]\[
dx = \frac{x}{3} du
\][/tex]
### Step 3: Rewriting the Integrand in terms of [tex]\( u \)[/tex]
Express the integrand using [tex]\( u \)[/tex]:
[tex]\[
u = 4 + 3 \ln (2x)
\][/tex]
Then,
[tex]\[
\ln(2x) = \frac{u - 4}{3}
\][/tex]
and the integrand becomes:
[tex]\[
\int \frac{\sqrt[3]{u}}{x} \cdot \frac{x}{3} \, du
\][/tex]
Simplifying the integrand:
[tex]\[
\int \frac{\sqrt[3]{u}}{3} \, du = \frac{1}{3} \int u^{1/3} \, du
\][/tex]
### Step 4: Integrate
Now we integrate [tex]\( u^{1/3} \)[/tex]:
[tex]\[
\frac{1}{3} \int u^{1/3} \, du = \frac{1}{3} \cdot \frac{3}{4} u^{4/3} + C = \frac{1}{4} u^{4/3} + C
\][/tex]
### Step 5: Substitute Back
Substitute back [tex]\( u = 4 + 3 \ln (2x) \)[/tex]:
[tex]\[
\frac{1}{4} (4 + 3 \ln (2x))^{4/3} + C
\][/tex]
### Final Answer
Thus, the finished integral is:
[tex]\[
\boxed{ \frac{1}{4}(4 + 3 \ln(2x))^{4/3} + C }
\][/tex]
