Let's solve both integrals step by step:

1. First Integral:

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

Solution: We use substitution to simplify the expression.

• Let $u = x^3 + 1$, so that $du = 3x^2 \, dx$.

Rewriting the integral: $\int 3x^2 \sqrt{x^3 + 1} \, dx = \int \sqrt{u} \, du$

Now integrate $\sqrt{u}$: $\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C$

Finally, substitute $u = x^3 + 1$ back: $\int 3x^2 \sqrt{x^3 + 1} \, dx = \frac{2}{3} (x^3 + 1)^{3/2} + C$

2. Second Integral:

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

Solution: We again use substitution.

• Let $u = x^2 + 1$, so $du = 2x \, dx$.

Rewriting the integral: $\int \frac{2x}{\sqrt{x^2 + 1}} \, dx = \int \frac{1}{\sqrt{u}} \, du$

Now integrate $\frac{1}{\sqrt{u}}$: $\int u^{-1/2} \, du = 2u^{1/2} + C$

Substitute $u = x^2 + 1$ back: $\int \frac{2x}{\sqrt{x^2 + 1}} \, dx = 2\sqrt{x^2 + 1} + C$

Final Answers:

1. $\int 3x^2 \sqrt{x^3 + 1} \, dx = \frac{2}{3} (x^3 + 1)^{3/2} + C$
2. $\int \frac{2x}{\sqrt{x^2 + 1}} \, dx = 2\sqrt{x^2 + 1} + C$

Let me know if you need further clarification or additional steps!

Related Questions:

1. How do you verify solutions to integrals using differentiation?
2. How do substitutions like $u = x^3 + 1$ work in integrals?
3. How can you apply integration by parts for more complex problems?
4. What are some other methods for solving square root integrals?
5. How do you solve definite integrals involving substitutions?

Tip:

Always double-check substitutions by differentiating the substituted function $u$ to ensure consistency.