Here are the solutions to the integrals provided in the image:

1. $\int e^{3x} \, dx$

The integral of $e^{kx}$ is $\frac{1}{k} e^{kx} + C$.

$\int e^{3x} \, dx = \frac{1}{3} e^{3x} + C$

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

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

$\int 2x e^{x^2 + 1} \, dx = \int e^u \, du = e^u + C = e^{x^2 + 1} + C$

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

Let $u = x + 2$. Then, $du = dx$, and the integral becomes:

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

4. $\int \sqrt{3x + 4} \, dx$

Let $u = 3x + 4$. Then, $du = 3 \, dx$, or $dx = \frac{du}{3}$. The integral becomes:

$\int \sqrt{3x + 4} \, dx = \int \sqrt{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \int u^{1/2} \, du$

$= \frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{9} (3x + 4)^{3/2} + C$

5. $\int x \cos(x^2) \, dx$

Let $u = x^2$. Then, $du = 2x \, dx$, or $\frac{du}{2} = x \, dx$. The integral becomes:

$\int x \cos(x^2) \, dx = \frac{1}{2} \int \cos(u) \, du = \frac{1}{2} \sin(u) + C = \frac{1}{2} \sin(x^2) + C$

Let me know if you would like further explanation for any step or additional details!

Related Questions:

1. How do substitution methods work in integrals?
2. What are the rules for integrating exponential functions?
3. How is trigonometric substitution applied in integration?
4. How can definite integrals be calculated for these problems?
5. What are some practical applications of these integrals in physics or engineering?

Tip:

Always check if a substitution can simplify your integral, especially when dealing with compositions of functions like $e^{x^2}$ or $\sqrt{x+2}$.