The problem requires solving the integral:

$\int \frac{x^4}{3 \sqrt{x-1}} \, dx$

using the substitution $u = \sqrt{x-1}$. Let's proceed step-by-step:

Step 1: Substitution $u = \sqrt{x-1}$

This implies: $u^2 = x - 1 \quad \text{or} \quad x = u^2 + 1.$

Now, differentiate $x = u^2 + 1$: $\frac{dx}{du} = 2u \quad \Rightarrow \quad dx = 2u \, du.$

Step 2: Express the integrand in terms of $u$

Replace $x$ and $dx$ in the integral: $\int \frac{x^4}{3 \sqrt{x-1}} \, dx \quad \Rightarrow \quad \int \frac{(u^2 + 1)^4}{3u} \cdot 2u \, du.$

Simplify: $\int \frac{(u^2 + 1)^4}{3u} \cdot 2u \, du = \frac{2}{3} \int (u^2 + 1)^4 \, du.$

Step 3: Expand $(u^2 + 1)^4$

Using the binomial theorem: $(u^2 + 1)^4 = \sum_{k=0}^4 \binom{4}{k} (u^2)^k (1)^{4-k}.$ The expanded form is: $(u^2 + 1)^4 = u^8 + 4u^6 + 6u^4 + 4u^2 + 1.$

Step 4: Integrate term-by-term

Substitute the expanded form into the integral: $\frac{2}{3} \int (u^2 + 1)^4 \, du = \frac{2}{3} \int (u^8 + 4u^6 + 6u^4 + 4u^2 + 1) \, du.$

Integrate each term: $\int u^8 \, du = \frac{u^9}{9}, \quad \int u^6 \, du = \frac{u^7}{7}, \quad \int u^4 \, du = \frac{u^5}{5}, \quad \int u^2 \, du = \frac{u^3}{3}, \quad \int 1 \, du = u.$

Thus: $\frac{2}{3} \int (u^8 + 4u^6 + 6u^4 + 4u^2 + 1) \, du = \frac{2}{3} \left( \frac{u^9}{9} + \frac{4u^7}{7} + \frac{6u^5}{5} + \frac{4u^3}{3} + u \right).$

Step 5: Simplify and substitute back $u = \sqrt{x-1}$

Multiply by $\frac{2}{3}$: $\frac{2}{3} \left( \frac{u^9}{9} + \frac{4u^7}{7} + \frac{6u^5}{5} + \frac{4u^3}{3} + u \right) = \frac{2u^9}{27} + \frac{8u^7}{21} + \frac{12u^5}{15} + \frac{8u^3}{9} + \frac{2u}{3}.$

Substitute back $u = \sqrt{x-1}$: $\frac{2}{3} \int \frac{x^4}{3 \sqrt{x-1}} \, dx = \frac{2 (\sqrt{x-1})^9}{27} + \frac{8 (\sqrt{x-1})^7}{21} + \frac{12 (\sqrt{x-1})^5}{15} + \frac{8 (\sqrt{x-1})^3}{9} + \frac{2 (\sqrt{x-1})}{3}.$

Would you like further clarification or help simplifying this result? Let me know!

Related Questions:

1. How do substitutions simplify integration problems involving square roots?
2. What is the general rule for applying the binomial theorem to expand polynomials?
3. How can one derive substitution formulas like $u = \sqrt{x-1}$?
4. Why is it important to return to the original variable after solving the integral?
5. Can similar substitutions help solve integrals with $\sqrt{x+a}$?

Tip:

Always check your substitution by differentiating your transformed variables to ensure correctness!