Sure, let's go through the partial fraction decomposition step by step for the rational expression
[tex]\[
\frac{x^2+3x+5}{x^3+x}.
\][/tex]
### Step 1: Factor the Denominator
First, we need to factor the denominator [tex]\(x^3 + x\)[/tex]. Notice that we can factor out an [tex]\(x\)[/tex], giving us:
[tex]\[
x^3 + x = x(x^2 + 1).
\][/tex]
So, the factored form of the denominator is [tex]\(x(x^2 + 1)\)[/tex].
### Step 2: Write the Partial Fraction Decomposition
Next, we express the rational function as the sum of partial fractions. Since the denominator factors as [tex]\(x(x^2 + 1)\)[/tex], we write:
[tex]\[
\frac{x^2 + 3x + 5}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}.
\][/tex]
Here, [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are constants that need to be determined.
### Step 3: Combine the Partial Fractions
To find [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex], we start by combining the partial fractions over a common denominator [tex]\(x(x^2 + 1)\)[/tex]:
[tex]\[
\frac{A}{x} + \frac{Bx + C}{x^2 + 1} = \frac{A(x^2 + 1) + (Bx + C)x}{x(x^2 + 1)}.
\][/tex]
Simplify the numerator:
[tex]\[
A(x^2 + 1) + (Bx + C)x = A x^2 + A + B x^2 x + C x = (A + B) x^2 + C x + A.
\][/tex]
### Step 4: Equate Numerators
Now, we equate the numerator of this expression to the numerator of the original rational function:
[tex]\[
x^2 + 3x + 5 = (A + B)x^2 + Cx + A.
\][/tex]
By comparing coefficients, we get the following system of equations:
1. [tex]\(A + B = 1\)[/tex]
2. [tex]\(C = 3\)[/tex]
3. [tex]\(A = 5\)[/tex]
### Step 5: Solve for A, B, and C
From the third equation, we have [tex]\(A = 5\)[/tex].
Substituting [tex]\(A = 5\)[/tex] into the first equation, we get:
[tex]\[
5 + B = 1 \implies B = -4.
\][/tex]
Thus, the coefficients are:
[tex]\[
A = 5, \quad B = -4, \quad C = 3.
\][/tex]
### Step 6: Write the Decomposed Form
So, the partial fraction decomposition of the rational expression is:
[tex]\[
\frac{x^2 + 3x + 5}{x(x^2 + 1)} = \frac{5}{x} + \frac{-4x + 3}{x^2 + 1}.
\][/tex]
### Step 7: Check the Result Algebraically
To verify, we need to recombine these fractions and check if we get back the original rational function:
[tex]\[
\frac{5}{x} + \frac{-4x + 3}{x^2 + 1} = \frac{5(x^2 + 1) + (-4x + 3)x}{x(x^2 + 1)}.
\][/tex]
Simplify the numerator:
[tex]\[
5(x^2 + 1) + (-4x + 3)x = 5x^2 + 5 - 4x^2 + 3x = x^2 + 3x + 5.
\][/tex]
So we have:
[tex]\[
\frac{x^2 + 3x + 5}{x(x^2 + 1)},
\][/tex]
which confirms our original rational function.
Thus, the correct partial fraction decomposition is
[tex]\[
\boxed{\frac{5}{x} + \frac{-4x + 3}{x^2 + 1}}.
\][/tex]
