The uploaded image contains integration problems that require solving by the substitution method. Here are the three problems:

1. $\int 2x \sqrt{x^2 + 3} \, dx$
2. $\int 5 \cos(3x) \, dx$
3. $\int x e^{1 - 3x^2} \, dx$

I'll solve each one using the substitution method step-by-step.

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

For this, we will use substitution.

Let: $u = x^2 + 3$ Then: $du = 2x \, dx$

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

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

Substitute back $u = x^2 + 3$: $\frac{2}{3} (x^2 + 3)^{3/2} + C$

b) $\int 5 \cos(3x) \, dx$

For this integral, we use substitution.

Let: $u = 3x$ Then: $du = 3 \, dx \quad \text{or} \quad dx = \frac{du}{3}$

Now, substitute in the integral: $\int 5 \cos(3x) \, dx = 5 \int \cos(u) \cdot \frac{du}{3} = \frac{5}{3} \int \cos(u) \, du$

Now, integrate: $\frac{5}{3} \sin(u) + C$

Substitute back $u = 3x$: $\frac{5}{3} \sin(3x) + C$

c) $\int x e^{1 - 3x^2} \, dx$

For this integral, we use substitution.

Let: $u = 1 - 3x^2$ Then: $du = -6x \, dx \quad \text{or} \quad x \, dx = -\frac{du}{6}$

Now, substitute in the integral: $\int x e^{1 - 3x^2} \, dx = \int e^u \cdot \left(-\frac{du}{6}\right) = -\frac{1}{6} \int e^u \, du$

Now, integrate: $-\frac{1}{6} e^u + C$

Substitute back $u = 1 - 3x^2$: $-\frac{1}{6} e^{1 - 3x^2} + C$

Let me know if you'd like further explanations or have any additional questions!

Here are some related questions you might find useful:

1. How do you determine the correct substitution for an integral?
2. What is the reverse chain rule, and how is it related to substitution?
3. Can you apply the substitution method to integrals with trigonometric functions?
4. What are some common mistakes when using substitution in integrals?
5. How does integration by parts differ from substitution?

Tip: Always double-check the derivative of your substitution to avoid errors during the process!