Let's solve these integrals one by one:

1. $\int x \sqrt{1+x^2} \, dx$

Use substitution: $u = 1+x^2$ → $du = 2x dx$ → $x dx = \frac{du}{2}$

The integral becomes: $\int x \sqrt{1+x^2} \, dx = \frac{1}{2} \int \sqrt{u} \, du$ Now integrate: $\frac{1}{2} \int u^{\frac{1}{2}} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} = \frac{1}{3} (1+x^2)^{\frac{3}{2}} + C$

2. $\int x \sqrt{x-1} \, dx$

Use substitution: $u = x-1$ → $x = u+1$ → $dx = du$

The integral becomes: $\int (u+1) \sqrt{u} \, du = \int \left( u^{\frac{3}{2}} + u^{\frac{1}{2}} \right) du$ Now integrate each term: $\int u^{\frac{3}{2}} du = \frac{2}{5} u^{\frac{5}{2}}, \quad \int u^{\frac{1}{2}} du = \frac{2}{3} u^{\frac{3}{2}}$ Therefore: $\int x \sqrt{x-1} \, dx = \frac{2}{5} (x-1)^{\frac{5}{2}} + \frac{2}{3} (x-1)^{\frac{3}{2}} + C$

3. $\int x^2 \sqrt{x-1} \, dx$

Use substitution: $u = x-1$ → $x = u+1$ → $dx = du$

The integral becomes: $\int (u+1)^2 \sqrt{u} \, du = \int \left( u^2 + 2u + 1 \right) u^{\frac{1}{2}} du$ Expand and integrate each term: $\int u^{\frac{5}{2}} du = \frac{2}{7} u^{\frac{7}{2}}, \quad \int 2 u^{\frac{3}{2}} du = \frac{4}{5} u^{\frac{5}{2}}, \quad \int u^{\frac{1}{2}} du = \frac{2}{3} u^{\frac{3}{2}}$ Therefore: $\int x^2 \sqrt{x-1} \, dx = \frac{2}{7} (x-1)^{\frac{7}{2}} + \frac{4}{5} (x-1)^{\frac{5}{2}} + \frac{2}{3} (x-1)^{\frac{3}{2}} + C$

Would you like more details on any step, or have any questions?

Related Questions:

1. How would the solutions change if different substitution methods were used?
2. What are some alternative integration techniques for these types of integrals?
3. How can these integrals be applied in physics or engineering contexts?
4. Can these integrals be evaluated using numerical methods?
5. What happens if the powers inside the square roots are different?

Tip:

For integrals involving square roots, substitution is often the key. Consider substituting the expression under the square root or related terms to simplify the integrand.