Integration Prep 1. Find $$ \int \frac{\sqrt{x}-3 x^{2}}{4 x^{3}} \mathrm{~d} x $$

Question

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To evaluate the integral:

$$
\int \frac{\sqrt{x} - 3x^{2}}{4x^{3}} \, dx
$$

**Step 1: Simplify the Integrand**

First, split the fraction and simplify each term:

$$
\frac{\sqrt{x} - 3x^{2}}{4x^{3}} = \frac{\sqrt{x}}{4x^{3}} - \frac{3x^{2}}{4x^{3}} = \frac{x^{1/2}}{4x^{3}} - \frac{3}{4x}
$$

Simplify the exponents:

$$
\frac{x^{1/2}}{4x^{3}} = \frac{1}{4}x^{1/2 - 3} = \frac{1}{4}x^{-5/2}
$$
$$
\frac{3x^{2}}{4x^{3}} = \frac{3}{4}x^{2-3} = \frac{3}{4}x^{-1}
$$

So the integral becomes:

$$
\frac{1}{4} \int x^{-5/2} \, dx - \frac{3}{4} \int x^{-1} \, dx
$$

**Step 2: Integrate Each Term**

Integrate each term separately:

1. **First Integral:**

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

2. **Second Integral:**

$$
\int x^{-1} \, dx = \ln|x| + C
$$

**Step 3: Combine the Results**

Substitute back into the expression:

$$
\frac{1}{4} \left(-\frac{2}{3}x^{-3/2}\right) - \frac{3}{4} \ln|x| + C = -\frac{1}{6}x^{-3/2} - \frac{3}{4}\ln|x| + C
$$

**Step 4: Express in Radical Form (Optional)**

If preferred, $$ x^{-3/2} $$ can be written as $$ \frac{1}{x^{3/2}} $$ or $$ \frac{1}{x \sqrt{x}} $$:

$$
-\frac{1}{6x^{3/2}} - \frac{3}{4}\ln|x| + C
$$

**Final Answer:**

$$
-\frac{1}{6\, x^{3/2}} \; - \; \frac{3}{4} \ln x \;+\; C
$$

where $$ C $$ is the constant of integration.
$$ \int \frac{\sqrt{x} - 3x^{2}}{4x^{3}} \, dx = -\frac{1}{6x^{3/2}} - \frac{3}{4}\ln|x| + C $$

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