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.