Sure, let's solve the integral [tex]\(\int \sqrt[3]{(2x - 1)^2} \, dx\)[/tex].
To solve this, we'll follow these steps:
### Step 1: Identify the integrand
The integrand is [tex]\(\sqrt[3]{(2x - 1)^2}\)[/tex]. This can be rewritten using fractional exponents:
[tex]\[ \sqrt[3]{(2x - 1)^2} = ((2x - 1)^2)^{1/3} = (2x - 1)^{2/3} \][/tex]
So, we need to integrate:
[tex]\[ \int (2x - 1)^{2/3} \, dx \][/tex]
### Step 2: Use substitution
Let's use the substitution method to simplify this integral. Let:
[tex]\[ u = 2x - 1 \][/tex]
Then, the differential [tex]\(du\)[/tex] is:
[tex]\[ du = 2 \, dx \implies dx = \frac{1}{2} \, du \][/tex]
### Step 3: Change the variable of integration
Using the substitution [tex]\(u = 2x - 1\)[/tex] and [tex]\(dx = \frac{1}{2} \, du\)[/tex], the integral becomes:
[tex]\[ \int (2x - 1)^{2/3} \, dx = \int u^{2/3} \cdot \frac{1}{2} \, du \][/tex]
[tex]\[ = \frac{1}{2} \int u^{2/3} \, du \][/tex]
### Step 4: Integrate with respect to [tex]\(u\)[/tex]
Now, we integrate [tex]\(u^{2/3}\)[/tex] with respect to [tex]\(u\)[/tex]:
[tex]\[ \int u^{2/3} \, du \][/tex]
To integrate this, we use the power rule for integration:
[tex]\[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C \][/tex]
Here, [tex]\(n = \frac{2}{3}\)[/tex]:
[tex]\[ \int u^{2/3} \, du = \frac{u^{2/3 + 1}}{2/3 + 1} + C = \frac{u^{5/3}}{5/3} + C = \frac{3u^{5/3}}{5} + C \][/tex]
### Step 5: Substitute back for [tex]\(u\)[/tex]
Recall that [tex]\(u = 2x - 1\)[/tex]:
[tex]\[ \frac{1}{2} \int u^{2/3} \, du = \frac{1}{2} \cdot \frac{3u^{5/3}}{5} + C \][/tex]
[tex]\[ = \frac{3}{10} u^{5/3} + C \][/tex]
Substituting back:
[tex]\[ \frac{3}{10} (2x - 1)^{5/3} + C \][/tex]
### Step 6: Combine the result
Including the constant of integration and simplifying, we get:
[tex]\[ \int \sqrt[3]{(2x - 1)^2} \, dx = \frac{3}{10} (2x - 1)^{5/3} + C \][/tex]
Therefore, the final answer is:
[tex]\[ \boxed{\frac{3}{10} (2x - 1)^{5/3} + C} \][/tex]
