`(i) int ((1+x)^(3))/(sqrt(x))dx " "(ii) int((1+x)^(3))/(x^(4)) dx`

Answer

Let's solve the given integrals step by step.

### Part (i): \(\int \frac{(1+x)^3}{\sqrt{x}} \, dx\)

1. **Expand \((1+x)^3\)**:
\[ (1+x)^3 = 1 + 3x + 3x^2 + x^3 \]

2. **Rewrite the integral**:
\[ \int \frac{(1+x)^3}{\sqrt{x}} \, dx = \int \frac{1 + 3x + 3x^2 + x^3}{\sqrt{x}} \, dx \]

3. **Express \(\sqrt{x}\) as \(x^{1/2}\)**:
\[ \int \frac{1 + 3x + 3x^2 + x^3}{x^{1/2}} \, dx \]

4. **Separate the integral**:
\[ = \int \left( x^{-1/2} + 3x^{1/2} + 3x^{3/2} + x^{5/2} \right) \, dx \]

5. **Integrate term by term**:
- For \(x^{-1/2}\):
\[ \int x^{-1/2} \, dx = 2x^{1/2} \]
- For \(3x^{1/2}\):
\[ \int 3x^{1/2} \, dx = 3 \cdot \frac{2}{3} x^{3/2} = 2x^{3/2} \]
- For \(3x^{3/2}\):
\[ \int 3x^{3/2} \, dx = 3 \cdot \frac{2}{5} x^{5/2} = \frac{6}{5} x^{5/2} \]
- For \(x^{5/2}\):
\[ \int x^{5/2} \, dx = \frac{2}{7} x^{7/2} \]

6. **Combine the results**:
\[ = 2x^{1/2} + 2x^{3/2} + \frac{6}{5} x^{5/2} + \frac{2}{7} x^{7/2} + C \]

### Final Answer for Part (i):
\[ \int \frac{(1+x)^3}{\sqrt{x}} \, dx = 2x^{1/2} + 2x^{3/2} + \frac{6}{5} x^{5/2} + \frac{2}{7} x^{7/2} + C \]

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### Part (ii): \(\int \frac{(1+x)^3}{x^4} \, dx\)

1. **Expand \((1+x)^3\)** (same as before):
\[ (1+x)^3 = 1 + 3x + 3x^2 + x^3 \]

2. **Rewrite the integral**:
\[ \int \frac{(1+x)^3}{x^4} \, dx = \int \frac{1 + 3x + 3x^2 + x^3}{x^4} \, dx \]

3. **Separate the integral**:
\[ = \int \left( x^{-4} + 3x^{-3} + 3x^{-2} + x^{-1} \right) \, dx \]

4. **Integrate term by term**:
- For \(x^{-4}\):
\[ \int x^{-4} \, dx = -\frac{1}{3} x^{-3} \]
- For \(3x^{-3}\):
\[ \int 3x^{-3} \, dx = -\frac{3}{2} x^{-2} \]
- For \(3x^{-2}\):
\[ \int 3x^{-2} \, dx = -3x^{-1} \]
- For \(x^{-1}\):
\[ \int x^{-1} \, dx = \ln |x| \]

5. **Combine the results**:
\[ = -\frac{1}{3} x^{-3} - \frac{3}{2} x^{-2} - 3x^{-1} + \ln |x| + C \]

### Final Answer for Part (ii):
\[ \int \frac{(1+x)^3}{x^4} \, dx = -\frac{1}{3} x^{-3} - \frac{3}{2} x^{-2} - 3x^{-1} + \ln |x| + C \]

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