It looks like you uploaded an image containing several math problems related to integration. Let's break them down one by one.

I’ll walk through each of the problems, so you can better understand how to approach solving them.

Problem 1:
$\int \left( 2 \sqrt{x} + 3 \right) \, dx$

This is an indefinite integral. To solve it, we can apply basic integration rules:

• For $2 \sqrt{x}$, rewrite as $2x^{1/2}$. The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$, so you’ll add 1 to the exponent and divide by the new exponent.
• For 3, the integral is simply $3x$.

Solution: $\int \left( 2x^{1/2} + 3 \right) \, dx = \frac{4}{3} x^{3/2} + 3x + C$

Problem 2:
Evaluate the definite integral: $\int_0^2 \frac{dx}{x^2 + 3}$

This is a rational function, and we can use a standard formula for the integral of $\frac{1}{x^2 + a^2}$, which is $\frac{1}{a} \arctan\left(\frac{x}{a}\right)$. For this problem, $a^2 = 3$, so $a = \sqrt{3}$.

Solution: $\int \frac{dx}{x^2 + 3} = \frac{1}{\sqrt{3}} \arctan\left(\frac{x}{\sqrt{3}}\right)$

Evaluating from 0 to 2: $\frac{1}{\sqrt{3}} \left[\arctan\left(\frac{2}{\sqrt{3}}\right) - \arctan(0)\right]$

This requires finding the arctangent of $2/\sqrt{3}$.

Problem 3:
Evaluate the definite integral: $\int_0^3 \sin(2t) \, dt$

The integral of $\sin(2t)$ is straightforward. Use the antiderivative of $\sin(kx)$, which is $-\frac{1}{k} \cos(kx)$.

Solution: $\int \sin(2t) \, dt = -\frac{1}{2} \cos(2t)$

Evaluating from 0 to 3: $-\frac{1}{2} \left[\cos(6) - \cos(0)\right]$

Problem 4:
Evaluate the definite integral: $\int_0^5 5 \sin(x) \cos(x) \, dx$

Use the identity $2 \sin(x) \cos(x) = \sin(2x)$ to simplify this. After applying that identity:

Solution: $\int_0^5 5 \sin(x) \cos(x) \, dx = \frac{5}{2} \int_0^5 \sin(2x) \, dx$

The integral of $\sin(2x)$ is $-\frac{1}{2} \cos(2x)$, so we evaluate this from 0 to 5.

Problem 5:
Evaluate the indefinite integral: $\int \frac{7x + 7}{x^2 + 1} \, dx$

This integral involves a rational function, and we can split the numerator: $\int \frac{7x}{x^2 + 1} \, dx + \int \frac{7}{x^2 + 1} \, dx$

For the first term, use the substitution $u = x^2 + 1$, and the second term is a standard arctangent integral.

Problem 6:
Evaluate the indefinite integral: $\int \frac{(\arcsin(x))^9}{\sqrt{1 - x^2}} \, dx$

This is a complicated integral, but recognizing that the term $\frac{1}{\sqrt{1 - x^2}}$ is the derivative of $\arcsin(x)$, this suggests a substitution approach.

Would you like to go through any of these problems in more detail, or do you need further clarification on any steps? Let me know!

Here are a few questions to dive deeper:

1. How does substitution work in solving integrals like $\int \frac{7x}{x^2 + 1} \, dx$?
2. Why do we use the identity $2 \sin(x) \cos(x) = \sin(2x)$ in integrals?
3. What are some common techniques for solving trigonometric integrals?
4. How does the constant of integration, $+ C$, appear in indefinite integrals?
5. What are the key differences between definite and indefinite integrals in terms of evaluation?

Tip: For definite integrals, always remember to substitute the limits after finding the antiderivative, not before!