Resolve into Partial Fractions
(vi) `(5x^(2)+2)/(x^(3)+x)`

Answer

To resolve the expression \(\frac{5x^2 + 2}{x^3 + x}\) into partial fractions, we will follow these steps:

### Step 1: Factor the Denominator
The denominator \(x^3 + x\) can be factored. We can take out \(x\) as a common factor:

\[ x^3 + x = x(x^2 + 1) \]

### Step 2: Set Up the Partial Fraction Decomposition
Since we have a linear factor \(x\) and an irreducible quadratic factor \(x^2 + 1\), we can express the fraction as:

\[ \frac{5x^2 + 2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \]

where \(A\), \(B\), and \(C\) are constants that we need to determine.

### Step 3: Combine the Right Side
To combine the right side into a single fraction, we find a common denominator:

\[ \frac{A}{x} + \frac{Bx + C}{x^2 + 1} = \frac{A(x^2 + 1) + (Bx + C)x}{x(x^2 + 1)} \]

This simplifies to:

\[ \frac{Ax^2 + A + Bx^2 + Cx}{x(x^2 + 1)} = \frac{(A + B)x^2 + Cx + A}{x(x^2 + 1)} \]

### Step 4: Set the Numerators Equal
Since the denominators are equal, we can set the numerators equal to each other:

\[ 5x^2 + 2 = (A + B)x^2 + Cx + A \]

### Step 5: Compare Coefficients
Now we compare coefficients from both sides:

1. Coefficient of \(x^2\): \(A + B = 5\)
2. Coefficient of \(x\): \(C = 0\)
3. Constant term: \(A = 2\)

### Step 6: Solve the System of Equations
From the equations we have:

- From \(A = 2\), we substitute into \(A + B = 5\):

\[ 2 + B = 5 \implies B = 3 \]

- We already have \(C = 0\).

### Step 7: Write the Partial Fraction Decomposition
Now substituting \(A\), \(B\), and \(C\) back into the partial fractions:

\[ \frac{5x^2 + 2}{x(x^2 + 1)} = \frac{2}{x} + \frac{3x + 0}{x^2 + 1} = \frac{2}{x} + \frac{3x}{x^2 + 1} \]

### Final Answer
Thus, the partial fraction decomposition of \(\frac{5x^2 + 2}{x^3 + x}\) is:

\[ \frac{2}{x} + \frac{3x}{x^2 + 1} \]