Let’s solve the two problems from the uploaded image.

1. Find $\int \frac{x + 5x^2 + 3}{\sqrt{x}} \, dx$:

We begin by rewriting the integral: $\int \frac{x + 5x^2 + 3}{\sqrt{x}} \, dx = \int \frac{x}{\sqrt{x}} \, dx + \int \frac{5x^2}{\sqrt{x}} \, dx + \int \frac{3}{\sqrt{x}} \, dx$

Each term is simplified using the property $\frac{x^n}{x^m} = x^{n-m}$: $\frac{x}{\sqrt{x}} = x^{1 - \frac{1}{2}} = x^{\frac{1}{2}}, \quad \frac{5x^2}{\sqrt{x}} = 5x^{2 - \frac{1}{2}} = 5x^{\frac{3}{2}}, \quad \frac{3}{\sqrt{x}} = 3x^{-\frac{1}{2}}$

So the integral becomes: $\int x^{\frac{1}{2}} \, dx + \int 5x^{\frac{3}{2}} \, dx + \int 3x^{-\frac{1}{2}} \, dx$

Now we integrate each term using the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$):

1. $\int x^{\frac{1}{2}} \, dx = \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}x^{\frac{3}{2}}$
2. $\int 5x^{\frac{3}{2}} \, dx = 5 \cdot \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} = 5 \cdot \frac{x^{\frac{5}{2}}}{\frac{5}{2}} = 5 \cdot \frac{2}{5}x^{\frac{5}{2}} = 2x^{\frac{5}{2}}$
3. $\int 3x^{-\frac{1}{2}} \, dx = 3 \cdot \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} = 3 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 3 \cdot 2x^{\frac{1}{2}} = 6x^{\frac{1}{2}}$

Adding them together, the result is: $\int \frac{x + 5x^2 + 3}{\sqrt{x}} \, dx = \frac{2}{3}x^{\frac{3}{2}} + 2x^{\frac{5}{2}} + 6x^{\frac{1}{2}} + C$

2. Find $\frac{d^3y}{dx^3}$ if $y = 2\sin(2x) + \cos(x)$:

Step 1: First derivative ($y'$):

$y = 2\sin(2x) + \cos(x)$

The derivatives of $\sin(2x)$ and $\cos(x)$ are:

• $\frac{d}{dx}\big[\sin(2x)\big] = 2\cos(2x)$ (by the chain rule),
• $\frac{d}{dx}\big[\cos(x)\big] = -\sin(x)$.

So: $y' = \frac{d}{dx}\big[2\sin(2x) + \cos(x)\big] = 2 \cdot 2\cos(2x) - \sin(x) = 4\cos(2x) - \sin(x)$

Step 2: Second derivative ($y''$):

$y' = 4\cos(2x) - \sin(x)$

Now differentiate term by term:

• $\frac{d}{dx}\big[4\cos(2x)\big] = 4 \cdot (-2\sin(2x)) = -8\sin(2x)$,
• $\frac{d}{dx}\big[-\sin(x)\big] = -\cos(x)$.

So: $y'' = -8\sin(2x) - \cos(x)$

Step 3: Third derivative ($y'''$):

$y'' = -8\sin(2x) - \cos(x)$

Now differentiate again:

• $\frac{d}{dx}\big[-8\sin(2x)\big] = -8 \cdot 2\cos(2x) = -16\cos(2x)$,
• $\frac{d}{dx}\big[-\cos(x)\big] = \sin(x)$.

So: $y''' = -16\cos(2x) + \sin(x)$

Final Answers:

1. $\int \frac{x + 5x^2 + 3}{\sqrt{x}} \, dx = \frac{2}{3}x^{\frac{3}{2}} + 2x^{\frac{5}{2}} + 6x^{\frac{1}{2}} + C$
2. $\frac{d^3y}{dx^3} = -16\cos(2x) + \sin(x)$

Do you want further clarification or more details?
Here are 5 related questions for further exploration:

1. How would you evaluate definite integrals for similar functions, such as $\int_1^4 \frac{x + 5x^2 + 3}{\sqrt{x}} \, dx$?
2. What is the importance of the chain rule in differentiation, as seen in the second problem?
3. How does the power rule for integration extend to functions with fractional exponents?
4. Can you explain how to verify the result of a third derivative using numerical methods?
5. How would the integral change if the numerator had higher powers of $x$, such as $x^3$?

Tip: Always rewrite integrals in terms of simpler powers before solving—it makes applying the power rule much easier!