Determine the following integrals: 9.1.1 \( \int m x^p \, dx \) where \( p \neq -1 \) 9.1.2 \( \int \frac{x^3 + x^2 - 2}{x^2 - 1} \, dx \) 9.2 The sketch below shows the shaded bounded area of the curve of the function defined by \( g(x) = \frac{4}{x}, \) where \( x > 0 - NSC Technical Mathematics - Question 9 - 2019 - Paper 1

For the integral, we will perform polynomial long division:

1. Polynomial Long Division: Divide ( x^3 + x^2 - 2 ) by ( x^2 - 1 ):

$$= x + 1 + \frac{-1}{x^2 - 1}.$$

2. Integration: Now we can integrate each term:

$$\int (x + 1) \, dx - \int \frac{1}{x^2 - 1} \, dx.$$

• For ( \int (x + 1) , dx ), the result is:

$$= \frac{x^2}{2} + x + C_1.$$

• For ( \int \frac{1}{x^2 - 1} , dx ), we can use partial fraction decomposition:

$$\frac{1}{x^2-1} = \frac{1/2}{x-1} - \frac{1/2}{x+1}.$$

This results in:

$$= \frac{1}{2} \ln |x-1| - \frac{1}{2} \ln |x+1| + C_2.$$

Thus, combining everything gives:

$$= \frac{x^2}{2} + x - \frac{1}{2} \ln |x^2 - 1| + C.$$

To find the shaded area, we need to evaluate the definite integral:

$$A = \int_{1}^{3.5} \frac{4}{x} \, dx.$$

1. Integration:

$$= 4 \int_{1}^{3.5} \frac{1}{x} \, dx = 4 [\ln |x|]_{1}^{3.5}.$$

2. Evaluating the bounds:

$$= 4 [\ln(3.5) - \ln(1)] = 4 \ln(3.5).$$

3. Calculate the area:

Using calculator:

$$A \approx 4 \cdot 1.25276 \approx 5.0110 \text{ square units}.$$

Thus, the shaded area is approximately:

$A \approx 5.0110 \text{ square units}.$