### Question 1:
Write each function $f(x)$ as a function composition $f(x) = g(u), u = u(x)$. Evaluate for $x = 2$

#### (a)
For $f(x) = \frac{2}{3x^2 - 1}$:

| Outside Function | Inside Function |
| --- | --- |
|$g(u) \hspace{3cm}$ | $u= \hspace{3cm}$|

<details> <summary> Example: </summary>

We want to express the function $f(x) = \frac{3}{2x^2 - 5}$ as a composition of functions $f(x) = g(u)$ and $u = u(x)$.

### Step 1: Define $u(x)$
Let $u(x) = 2x^2 - 5$. This is the expression inside the denominator.

### Step 2: Define $g(u)$
Now, express $f(x)$ in terms of $u$. Since $f(x) = \frac{3}{2x^2 - 5}$, we can write:
$$
g(u) = \frac{3}{u}
$$

Thus, we have:
$$
f(x) = g(u(x)) = g(2x^2 - 5) = \frac{3}{2x^2 - 5}
$$

| Outside Function | Inside Function |
| --- | --- |
| $g(u) = \dfrac{3}{u}$ | $u = 2x^2 - 5$ |

### Step 3: Evaluate at $x = 2$
Now, substitute $x = 2$ into $u(x)$:
$$
u(2) = 2(2)^2 - 5 = 2(4) - 5 = 8 - 5 = 3
$$
Now, evaluate $g(u)$ at $u = 3$:
$$
g(3) = \frac{3}{3} = 1
$$

Thus, $f(2) = 1$.

</details>

#### (b)
For $f(x) = -\frac{3}{(5 - 2x)^3}$:
| Outside Function | Inside Function |
| --- | --- |
|$g(u) \hspace{3cm}$ | $u=\hspace{3cm}$|

<details> <summary> Example: </summary>

For $f(x) = -\frac{4}{(5 - 3x)^2}$:

We want to express the function $f(x) = -\frac{4}{(5 - 3x)^2}$ as a composition of functions $f(x) = g(u)$ and $u = u(x)$.

### Step 1: Define $u(x)$
Let $u(x) = 5 - 3x$. This is the expression inside the parentheses.

### Step 2: Define $g(u)$
Now, express $f(x)$ in terms of $u$. Since $f(x) = -\frac{4}{(5 - 3x)^2}$, we can write:
$$
g(u) = -\frac{4}{u^2}
$$

Thus, we have:
$$
f(x) = g(u(x)) = g(5 - 3x) = -\frac{4}{(5 - 3x)^2}
$$

| Outside Function | Inside Function |
| --- | --- |
| $g(u) = -\frac{4}{u^2}$ | $u = 5 - 3x$ |

### Step 3: Evaluate at $x = 2$
Now, substitute $x = 2$ into $u(x)$:
$$
u(2) = 5 - 3(2) = 5 - 6 = -1
$$
Now, evaluate $g(u)$ at $u = -1$:
$$
g(-1) = -\frac{4}{(-1)^2} = -\frac{4}{1} = -4
$$

Thus, $f(2) = -4$.

</details>

#### (c)
For $f(x) = 2\ln(3x^4 - 2x^2)$:
| Outside Function | Inside Function |
| --- | --- |
|$g(u) \hspace{3cm}$ | $u=\hspace{3cm}$|

<details> <summary> Example: </summary>

We want to express the function $f(x) = 3\ln(2x^3 - 4x)$ as a composition of functions $f(x) = g(u)$ and $u = u(x)$.

### Step 1: Define $u(x)$
Let $u(x) = 2x^3 - 4x$. This is the expression inside the logarithm.

### Step 2: Define $g(u)$
Now, express $f(x)$ in terms of $u$. Since $f(x) = 3\ln(2x^3 - 4x)$, we can write:
$$
g(u) = 3\ln(u)
$$

Thus, we have:
$$
f(x) = g(u(x)) = g(2x^3 - 4x) = 3\ln(2x^3 - 4x)
$$

| Outside Function | Inside Function |
| --- | --- |
| $g(u) = 3\ln(u)$ | $u = 2x^3 - 4x$ |

### Step 3: Evaluate at $x = 2$
Now, substitute $x = 2$ into $u(x)$:
$$
u(2) = 2(2)^3 - 4(2) = 2(8) - 8 = 16 - 8 = 8
$$
Now, evaluate $g(u)$ at $u = 8$:
$$
g(8) = 3\ln(8)
$$

Thus, $f(2) = 3\ln(8)$.

</details>

---

### Question 2:
Find the following derivatives

#### (a)
For $f(x) = xe^x$:
$$
f'(x) =
$$

<details> <summary> Example: </summary>

We are tasked with finding the derivative of the function $f(x) = x^2 e^x$. We will use the product rule to find the derivative.

### Derivative Table

| | First Function | Second Function |
| --- | --- | --- |
| **Original** | $x^2$ | $e^x$ |
| **Derivative** | $2x$ | $e^x$ |

Using the product rule:
$$
f'(x) = (2x)(e^x) + (x^2)(e^x)
$$

### Step 2: Simplify
Factor out the common term $xe^x$:
$$
f'(x) = xe^x(2 + x)
$$

Thus, the derivative of $f(x) = x^2 e^x$ is:
$$
f'(x) = xe^x(2 + x)
$$

</details>

#### (b)
For $g(x) = x\ln(x)$:
$$
g'(x) =
$$

<details> <summary> Example: </summary>

We are tasked with finding the derivative of the function $f(x) = x^2 \ln x$. We will use the product rule to find the derivative.

### Derivative Table

| | First Function | Second Function |
| --- | --- | --- |
| **Original** | $x^2$ | $\ln x$ |
| **Derivative** | $2x$ | $\frac{1}{x}$ |

Using the product rule:
$$
f'(x) = (2x)(\ln x) + (x^2)\left(\frac{1}{x}\right)
$$

### Step 2: Simplify
Simplify the second term:
$$
f'(x) = 2x \ln x + x
$$

Thus, the derivative of $f(x) = x^2 \ln x$ is:
$$
f'(x) = 2x \ln x + x
$$

</details>

#### (c)
For $h(x) = \frac{x}{x - 1}$:
$$
h'(x) =
$$

<details> <summary> Example: </summary>

We are tasked with finding the derivative of the function $h(x) = \frac{x}{x - 2}$. We will use the quotient rule to find the derivative.

### Quotient Rule
The quotient rule states that for two functions $u(x)$ and $v(x)$, the derivative of their quotient is:
$$
h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
$$
Let:
- $u(x) = x$
- $v(x) = x - 2$

### Derivative Table

| | Numerator (First Function) | Denominator (Second Function) |
| --- | --- | --- |
| **Original** | $x$ | $x - 2$ |
| **Derivative** | $1$ | $1$ |

Using the quotient rule:
$$
h'(x) = \frac{(1)(x - 2) - (x)(1)}{(x - 2)^2}
$$

### Step 2: Simplify
Simplify the expression:
$$
h'(x) = \frac{x - 2 - x}{(x - 2)^2} = \frac{-2}{(x - 2)^2}
$$

Thus, the derivative of $h(x) = \frac{x}{x - 2}$ is:
$$
h'(x) = \frac{-2}{(x - 2)^2}
$$

</details>

### Question 3:
Find each indefinite integral

#### (a)
$$
\int (3x + 5)^5 (3) \, dx =
$$

<details> <summary> Example: </summary>

$$
\int (4x + 3)^2 (4) \, dx =
$$
The simplest way to approach this integral using $u$-substitution is to solve for $dx$ so that the $4$ cancels out.

### Step 1: Let $u = 4x + 3$
Differentiate both sides to find $dx$:
$$
du = 4 \, dx
$$
Solve for $dx$:
$$
dx = \frac{du}{4}
$$

### Step 2: Substitute into the integral
Now substitute $u = 4x + 3$ and $dx = \frac{du}{4}$ into the integral:
$$
\int (4x + 3)^2 (4) \, dx = \int u^2 (4) \left( \frac{du}{4} \right)
$$

The $4$ cancels out:
$$
\int u^2 \, du
$$

### Step 3: Integrate
Now integrate $u^2$:
$$
\int u^2 \, du = \frac{u^3}{3} + C
$$

### Step 4: Substitute $u = 4x + 3$ back
Substitute $u = 4x + 3$ back into the result:
$$
\frac{(4x + 3)^3}{3} + C
$$

Thus, the integral is:
$$
\int (4x + 3)^2 (4) \, dx = \frac{(4x + 3)^3}{3} + C
$$

</details>

#### (b)
$$
\int \frac{1}{1 + x^2} (2x) \, dx =
$$

<details> <summary> Example: </summary>

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

The simplest way to approach this integral using $u$-substitution is to solve for $dx$ so that the $3x^2$ cancels out.

### Step 1: Let $u = 1 + x^3$
Differentiate both sides to find $dx$:
$$
du = 3x^2 \, dx
$$
Solve for $dx$:
$$
dx = \frac{du}{3x^2}
$$

### Step 2: Substitute into the integral
Now substitute $u = 1 + x^3$ and $dx = \frac{du}{3x^2}$ into the integral:
$$
\int \frac{1}{1 + x^3} (3x^2) \, dx = \int \frac{1}{u} (3x^2) \left( \frac{du}{3x^2} \right)
$$

The $3x^2$ cancels out:
$$
\int \frac{1}{u} \, du
$$

### Step 3: Integrate
Now integrate $\frac{1}{u}$:
$$
\int \frac{1}{u} \, du = \ln|u| + C
$$

### Step 4: Substitute $u = 1 + x^3$ back
Substitute $u = 1 + x^3$ back into the result:
$$
\ln|1 + x^3| + C
$$

Thus, the integral is:
$$
\int \frac{1}{1 + x^3} (3x^2) \, dx = \ln|1 + x^3| + C
$$

</details>

#### (c)
$$
\int \frac{t^2}{(t^3 - 2)^5} \, dt =
$$

<details> <summary> Example: </summary>

$$
\int \frac{t^3}{(t^4 - 1)^5} \, dt =
$$

The simplest way to approach this integral using $u$-substitution is to find an expression for $u$ and solve for $dt$ so that the expression simplifies.

### Step 1: Let $u = t^4 - 1$
Differentiate both sides to find $dt$:
$$
du = 4t^3 \, dt
$$
Solve for $dt$:
$$
dt = \frac{du}{4t^3}
$$

### Step 2: Substitute into the integral
Now substitute $u = t^4 - 1$ and $dt = \frac{du}{4t^3}$ into the integral:
$$
\int \frac{t^3}{(t^4 - 1)^5} \, dt = \int \frac{t^3}{u^5} \left( \frac{du}{4t^3} \right)
$$

The $t^3$ cancels out:
$$
\int \frac{1}{4u^5} \, du=\dfrac{1}{4} \int u^{-5} \, du
$$

### Step 3: Integrate
Now integrate $\dfrac{1}{4} \int u^{-5} \, du$:
$$
\dfrac{1}{4} \int u^{-5} \, du = \frac{1}{4} \cdot \frac{u^{-4}}{-4} + C = -\frac{1}{16u^4} + C
$$

### Step 4: Substitute $u = t^4 - 1$ back
Substitute $u = t^4 - 1$ back into the result:
$$
-\frac{1}{16(t^4 - 1)^4} + C
$$

Thus, the integral is:
$$
\int \frac{t^3}{(t^4 - 1)^5} \, dt = -\frac{1}{16(t^4 - 1)^4} + C
$$

</details>

#### (d)
$$
\int \frac{x^3}{\sqrt{3x^4 + 1}} \, dx =
$$

<details> <summary> Example: </summary>

$$
\int \frac{x^4}{\sqrt{2x^5 + 3}} \, dx =
$$

We are tasked with finding the integral of the function $\int \frac{x^4}{\sqrt{2x^5 + 3}} \, dx$. We will use $u$-substitution and simplify the integrand before integrating.

### Step 1: Let $u = 2x^5 + 3$
Differentiate both sides to find $dx$:
$$
du = 10x^4 \, dx
$$
Solve for $dx$:
$$
dx = \frac{du}{10x^4}
$$

### Step 2: Substitute into the integral
Substitute $u = 2x^5 + 3$ and $dx = \frac{du}{10x^4}$ into the integral:
$$
\int \frac{x^4}{\sqrt{2x^5 + 3}} \, dx = \int \frac{x^4}{\sqrt{u}} \cdot \frac{du}{10x^4}
$$

The $x^4$ cancels out:
$$
\frac{1}{10} \int \frac{1}{\sqrt{u}} \, du
$$

### Step 3: Simplify the integrand
We can rewrite the integrand as $u^{-1/2}$:
$$
\frac{1}{10} \int u^{-1/2} \, du
$$

### Step 4: Integrate
Now integrate $u^{-1/2}$:
$$
\frac{1}{10} \cdot \frac{u^{1/2}}{1/2} = \frac{1}{10} \cdot 2u^{1/2} = \frac{1}{5} u^{1/2} + C
$$

### Step 5: Substitute $u = 2x^5 + 3$ back
Substitute $u = 2x^5 + 3$ back into the result:
$$
\frac{1}{5} \sqrt{2x^5 + 3} + C
$$

Thus, the integral is:
$$
\int \frac{x^4}{\sqrt{2x^5 + 3}} \, dx = \frac{1}{5} \sqrt{2x^5 + 3} + C
$$

</details>