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hazme los ejercicos de manera explicita y con integración directa 

Let's solve each of the integrals one by one using direct integration. I'll go through them in detail:

### 1. Integral \(\int (x^2 + 1)(\sqrt{x^3 - x}) \, dx\)
Here, the integration looks a bit tricky because of the square root. However, we can try a substitution method.

Let \( u = x^3 - x \).

Then,
\[
du = (3x^2 - 1) \, dx.
\]

Rewriting the integral in terms of \( u \):

\[
\int (x^2 + 1) \sqrt{x^3 - x} \, dx = \int \left( \frac{du}{3} \right) \sqrt{u} \cdot \left(\frac{1}{x^2} + 1\right) = \int \frac{1}{3} u^{1/2} du 
\]

Therefore, the integral becomes:

\[
\int \frac{1}{3} u^{1/2} \, du = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{9} \left(x^3 - x \right)^{3/2} + C
\]

### 2. Integral \(\int \frac{(x^2 + 1)(x^2 - 1)}{\sqrt{x}} \, dx\)
Expand the expression in the numerator:

\[
\int \frac{(x^4 - 1)}{\sqrt{x}} \, dx.
\]

Now distribute the \(\sqrt{x}\) to each term:

\[
\int \left( x^{\frac{7}{2}} - x^{\frac{-1}{2}} \right) \, dx.
\]

Integrate each term directly:

\[
\int x^{\frac{7}{2}} \, dx - \int x^{-\frac{1}{2}} \, dx = \frac{2}{9}x^{\frac{9}{2}} - 2x^{\frac{1}{2}} + C.
\]

### 3. Integral \(\int \left(\frac{3}{\sqrt{t}} - \frac{\sqrt{t}}{4} \right) \, dt\)
This integral can be split up and solved term by term:

\[
\int \frac{3}{\sqrt{t}} \, dt - \frac{1}{4} \int \sqrt{t} \, dt.
\]

Integrating each term:

\[
3 \cdot 2 \cdot t^{\frac{1}{2}} - \frac{1}{4} \cdot \frac{2}{3} \cdot t^{\frac{3}{2}} + C = 6t^{\frac{1}{2}} - \frac{1}{6} t^{\frac{3}{2}} + C.
\]

### 4. Integral \(\int \frac{-x^2}{x^2 + 5} \, dx\)
This can be solved using a simple substitution. Let \( u = x^2 + 5 \).

Then,
\[
du = 2x \, dx.
\]

The integral becomes:

\[
-\frac{1}{2} \int \frac{u-5}{u} du = -\frac{1}{2} \left[ \int du - 5\int \frac{1}{u} \, du \right].
\]

This simplifies to:

\[
-\frac{1}{2} \left[ u - 5\ln |u| \right] + C.
\]

Substituting back \( u = x^2 + 5 \):

\[
-\frac{1}{2} \left[ x^2 + 5 - 5\ln |x^2 + 5| \right] + C.
\]

### 5. Integral \(\int \cot^2(\theta) \, d\theta\)
We know that \( \cot^2(\theta) = \csc^2(\theta) - 1 \).

So the integral becomes:

\[
\int \csc^2(\theta) \, d\theta - \int 1 \, d\theta.
\]

The integral of \( \csc^2(\theta) \) is \( -\cot(\theta) \), and the integral of \( 1 \) is \( \theta \):

\[
-\cot(\theta) - \theta + C.
\]

---

### Summary of Solutions:

1. \(\frac{2}{9} (x^3 - x)^{3/2} + C\)
2. \( \frac{2}{9} x^{\frac{9}{2}} - 2x^{\frac{1}{2}} + C \)
3. \( 6t^{\frac{1}{2}} - \frac{1}{6} t^{\frac{3}{2}} + C \)
4. \( -\frac{1}{2} \left[ x^2 + 5 - 5\ln |x^2 + 5| \right] + C \)
5. \( -\cot(\theta) - \theta + C \)

Would you like any further details or have any questions? 

Here are some related questions to explore further:
1. How can substitution be used to simplify integrals?
2. What are common integrals involving trigonometric functions?
3. How can partial fractions help in integration?
4. What are the basic integration formulas that are commonly used?
5. How do you handle integrals with square roots?

**Tip:** When dealing with complex integrals, substitution can be a powerful tool to simplify the problem.

Explore detailed solutions for tricky integrals using direct integration methods. Learn how to solve integrals involving substitution, trigonometric functions, and more. Suitable for undergraduate level mathematics.

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