To solve the integral \( \int 2x \sqrt{x-1} \, dx \), we will use substitution. Here's the step-by-step process:

### Step 1: Substitution
Let:
\[
u = x - 1
\]
Then:
\[
du = dx \quad \text{and} \quad x = u + 1
\]

### Step 2: Substituting in terms of \( u \)
Now, substitute \( x = u + 1 \) and \( \sqrt{x-1} = \sqrt{u} \) into the integral:
\[
\int 2x \sqrt{x-1} \, dx = \int 2(u + 1) \sqrt{u} \, du
\]

### Step 3: Expand the integrand
Expand the expression:
\[
\int 2(u + 1) \sqrt{u} \, du = \int 2(u \sqrt{u} + \sqrt{u}) \, du
\]
Simplifying each term:
\[
= \int 2u^{3/2} \, du + \int 2u^{1/2} \, du
\]

### Step 4: Integrate each term
Now, integrate each term separately.

For \( \int 2u^{3/2} \, du \):
\[
\int 2u^{3/2} \, du = \frac{2}{5} u^{5/2}
\]

For \( \int 2u^{1/2} \, du \):
\[
\int 2u^{1/2} \, du = \frac{4}{3} u^{3/2}
\]

### Step 5: Combine the results
Now combine the results:
\[
\frac{2}{5} u^{5/2} + \frac{4}{3} u^{3/2} + C
\]

### Step 6: Substitute \( u = x - 1 \) back
Now, substitute \( u = x - 1 \) back into the expression:
\[
\frac{2}{5} (x-1)^{5/2} + \frac{4}{3} (x-1)^{3/2} + C
\]

Thus, the solution to the integral is:
\[
\int 2x \sqrt{x-1} \, dx = \frac{2}{5} (x-1)^{5/2} + \frac{4}{3} (x-1)^{3/2} + C
\]

Would you like any further clarification or more details?

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Here are five related questions:
1. How do we choose a suitable substitution for integration?
2. Can you solve the integral \( \int x \sqrt{x+2} \, dx \)?
3. How does integration by parts differ from substitution?
4. What are some common integrals involving square roots?
5. Can the chain rule be used for integration, similar to differentiation?

**Tip:** Always check if substitution simplifies the function before trying other methods like integration by parts.

Learn how to solve the integral ∫ 2x √(x-1) dx step-by-step using substitution. This solution covers key calculus concepts like substitution and integration of radicals. Suitable for advanced high school or college-level calculus students.

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