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Can you do questions 29,31,33,35 and show all the step/work please

I can now see the integrals from the image you provided. Let's go through them one by one, showing all steps and solutions.

### 29. \( \int 3e^{-0.2x} \, dx \)

#### Solution:
This is an exponential integral of the form \( \int e^{ax} \, dx \).

1. We start by factoring out the constant:
   \[
   \int 3e^{-0.2x} \, dx = 3 \int e^{-0.2x} \, dx
   \]

2. Now, integrate \( e^{-0.2x} \):
   The integral of \( e^{ax} \) is \( \frac{e^{ax}}{a} \), where \( a = -0.2 \).

   \[
   3 \int e^{-0.2x} \, dx = 3 \left( \frac{e^{-0.2x}}{-0.2} \right) = -15e^{-0.2x}
   \]

3. Add the constant of integration \( C \):
   \[
   \boxed{-15e^{-0.2x} + C}
   \]

---

### 31. \( \int \left( \frac{-3}{x} + 4e^{-0.4x} + e^{0.1} \right) dx \)

#### Solution:
We can break this into three simpler integrals:
\[
\int \left( \frac{-3}{x} \, dx \right) + \int 4e^{-0.4x} \, dx + \int e^{0.1} \, dx
\]

1. For \( \int \frac{-3}{x} \, dx \), we use the fact that \( \int \frac{1}{x} \, dx = \ln|x| \):
   \[
   \int \frac{-3}{x} \, dx = -3 \ln|x|
   \]

2. For \( \int 4e^{-0.4x} \, dx \), we treat this as an exponential integral:
   \[
   \int 4e^{-0.4x} \, dx = 4 \left( \frac{e^{-0.4x}}{-0.4} \right) = -10e^{-0.4x}
   \]

3. For \( \int e^{0.1} \, dx \), notice that \( e^{0.1} \) is a constant:
   \[
   \int e^{0.1} \, dx = e^{0.1}x
   \]

Thus, putting it all together:
\[
\boxed{-3 \ln|x| - 10e^{-0.4x} + e^{0.1}x + C}
\]

---

### 33. \( \int \frac{1 + 2t^3}{4t} \, dt \)

#### Solution:
We can split this integral:
\[
\int \frac{1}{4t} \, dt + \int \frac{2t^3}{4t} \, dt
\]

1. For \( \int \frac{1}{4t} \, dt \):
   \[
   \int \frac{1}{4t} \, dt = \frac{1}{4} \ln|t|
   \]

2. Simplify \( \frac{2t^3}{4t} = \frac{t^2}{2} \), and then integrate:
   \[
   \int \frac{t^2}{2} \, dt = \frac{1}{2} \int t^2 \, dt = \frac{1}{2} \left( \frac{t^3}{3} \right) = \frac{t^3}{6}
   \]

Thus, the integral is:
\[
\boxed{\frac{1}{4} \ln|t| + \frac{t^3}{6} + C}
\]

---

### 35. \( \int \left( e^{2u} + 4u \right) \, du \)

#### Solution:
This is a sum of two integrals:
\[
\int e^{2u} \, du + \int 4u \, du
\]

1. For \( \int e^{2u} \, du \), we apply the rule for exponential functions:
   \[
   \int e^{2u} \, du = \frac{e^{2u}}{2}
   \]

2. For \( \int 4u \, du \), we apply the power rule \( \int u^n \, du = \frac{u^{n+1}}{n+1} \):
   \[
   \int 4u \, du = 4 \left( \frac{u^2}{2} \right) = 2u^2
   \]

Thus, the integral becomes:
\[
\boxed{\frac{e^{2u}}{2} + 2u^2 + C}
\]

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