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Mostre os calculos e a resposta das questões da imagem

Let's go through each question one by one, calculating the indefinite integrals as requested.

### Questão 01
**a) \(\int \left(s^3 - 2s + 3\right) \, ds\)**

To solve this, we can integrate each term separately:

\[
\int s^3 \, ds = \frac{s^4}{4}
\]
\[
\int -2s \, ds = -s^2
\]
\[
\int 3 \, ds = 3s
\]

So, the integral is:
\[
\int \left(s^3 - 2s + 3\right) \, ds = \frac{s^4}{4} - s^2 + 3s + C
\]
where \(C\) is the constant of integration.

**b) \(\int \left(9t^2 + \frac{1}{\sqrt{t^3}}\right) \, dt\)**

Here, we also integrate term by term:

\[
\int 9t^2 \, dt = 3t^3
\]

For the second term, rewrite it as \( t^{-\frac{3}{2}} \):

\[
\int t^{-\frac{3}{2}} \, dt = \frac{t^{-\frac{1}{2}}}{-\frac{1}{2}} = -2t^{-\frac{1}{2}} = -\frac{2}{\sqrt{t}}
\]

Thus, the integral is:
\[
\int \left(9t^2 + \frac{1}{\sqrt{t^3}}\right) \, dt = 3t^3 - \frac{2}{\sqrt{t}} + C
\]

### Questão 02
**a) \(\int \left(3x - \frac{1}{x^7} + \frac{1}{3x}\right) \, dx\)**

This is another case of integrating each term separately:

\[
\int 3x \, dx = \frac{3x^2}{2}
\]

\[
\int -\frac{1}{x^7} \, dx = -\frac{x^{-6}}{-6} = \frac{1}{6x^6}
\]

\[
\int \frac{1}{3x} \, dx = \frac{1}{3} \ln |x|
\]

So, the integral is:
\[
\int \left(3x - \frac{1}{x^7} + \frac{1}{3x}\right) \, dx = \frac{3x^2}{2} + \frac{1}{6x^6} + \frac{1}{3} \ln |x| + C
\]

**b) \(\int \frac{d\theta}{\sin^2\theta}\)**

This integral can be rewritten using the trigonometric identity \(\csc^2\theta = \frac{1}{\sin^2\theta}\):

\[
\int \csc^2\theta \, d\theta = -\cot\theta + C
\]

### Questão 03
**\(\int v(v^2 + 2)^2 \, dv\)**

This requires expanding the polynomial and then integrating:

Expand \((v^2 + 2)^2\):
\[
(v^2 + 2)^2 = v^4 + 4v^2 + 4
\]

Now, multiply by \(v\) and integrate:
\[
v(v^4 + 4v^2 + 4) = v^5 + 4v^3 + 4v
\]

The integral becomes:
\[
\int \left(v^5 + 4v^3 + 4v\right) \, dv = \frac{v^6}{6} + v^4 + 2v^2 + C
\]

### Questão 04
**\(\int \frac{2t^2 + t^2\sqrt{t} - 1}{t^2} \, dt\)**

First, simplify the integrand:

\[
\frac{2t^2}{t^2} + \frac{t^2\sqrt{t}}{t^2} - \frac{1}{t^2} = 2 + t^{-\frac{1}{2}} - t^{-2}
\]

Now, integrate each term:

\[
\int 2 \, dt = 2t
\]
\[
\int t^{-\frac{1}{2}} \, dt = 2t^{\frac{1}{2}}
\]
\[
\int -t^{-2} \, dt = \frac{t^{-1}}{1} = -\frac{1}{t}
\]

Thus, the integral is:
\[
\int \frac{2t^2 + t^2\sqrt{t} - 1}{t^2} \, dt = 2t + 2\sqrt{t} - \frac{1}{t} + C
\]

---

Would you like more details on any specific step or further questions?

### Related Questions:
1. How do you compute the integral of a trigonometric function with a power like \(\sin^3 x\)?
2. Can you show how to solve an integral involving an exponential function, like \(\int e^{2x} \, dx\)?
3. What is the integral of a logarithmic function, for example, \(\int \ln(x) \, dx\)?
4. How would you approach integrating by parts for the function \(\int x e^x \, dx\)?
5. Can you explain how to handle integrals that involve partial fractions?

### Tip:
When integrating, always consider simplifying the integrand first if possible. Simplifications can make the integration process much easier and more straightforward.

Explore como resolver questões de cálculo integral com passos detalhados e respostas completas. Inclui integração por partes, fórmulas de integrais indefinidas e teorema fundamental do cálculo. Adequado para estudantes de nível universitário.

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