Answer :
Let's tackle the problem step-by-step.
### Step 1: Partial Fraction Decomposition
We need to decompose the integrand into partial fractions.
The integrand is:
[tex]\[ \frac{x^2 - x + 6}{x^3 + 3x} \][/tex]
First, we factor the denominator:
[tex]\[ x^3 + 3x = x(x^2 + 3) \][/tex]
So, we can express the integrand as:
[tex]\[ \frac{x^2 - x + 6}{x(x^2 + 3)} \][/tex]
We assume a partial fraction decomposition of the form:
[tex]\[ \frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3} \][/tex]
Next, we multiply both sides by the denominator [tex]\( x(x^2 + 3) \)[/tex] to clear the fractions:
[tex]\[ x^2 - x + 6 = A(x^2 + 3) + (Bx + C)x \][/tex]
Now we expand and combine like terms:
[tex]\[ x^2 - x + 6 = Ax^2 + 3A + Bx^2 + Cx \][/tex]
[tex]\[ x^2 - x + 6 = (A + B)x^2 + Cx + 3A \][/tex]
We equate the coefficients of the corresponding powers of [tex]\(x\)[/tex] on both sides:
1. For [tex]\(x^2\)[/tex]: [tex]\( A + B = 1 \)[/tex]
2. For [tex]\(x\)[/tex]: [tex]\( C = -1 \)[/tex]
3. For the constant term: [tex]\( 3A = 6 \)[/tex]
From equation (3):
[tex]\[ 3A = 6 \Rightarrow A = 2 \][/tex]
Substitute [tex]\( A = 2 \)[/tex] into equation (1):
[tex]\[ 2 + B = 1 \Rightarrow B = -1 \][/tex]
So, the partial fraction decomposition is:
[tex]\[ \frac{2}{x} + \frac{(-1)x - 1}{x^2 + 3} \][/tex]
[tex]\[ = \frac{2}{x} - \frac{x + 1}{x^2 + 3} \][/tex]
### Step 2: Integrate the Partial Fractions
Now, we integrate each term separately:
[tex]\[ \int \left( \frac{2}{x} - \frac{x + 1}{x^2 + 3} \right) dx \][/tex]
First, integrate [tex]\( \frac{2}{x} \)[/tex]:
[tex]\[ \int \frac{2}{x} dx = 2 \ln |x| \][/tex]
Next, integrate [tex]\( \frac{x + 1}{x^2 + 3} \)[/tex]. Break it into two integrals:
[tex]\[ \int \frac{x + 1}{x^2 + 3} dx = \int \frac{x}{x^2 + 3} dx + \int \frac{1}{x^2 + 3} dx \][/tex]
For [tex]\( \int \frac{x}{x^2 + 3} dx \)[/tex]:
Use the substitution [tex]\( u = x^2 + 3 \)[/tex], then [tex]\( du = 2x \, dx \)[/tex]:
[tex]\[ \int \frac{x}{x^2 + 3} dx = \frac{1}{2} \int \frac{2x}{x^2 + 3} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|x^2 + 3| \][/tex]
For [tex]\( \int \frac{1}{x^2 + 3} dx \)[/tex]:
Recall the integral of [tex]\( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) \)[/tex]:
[tex]\[ \int \frac{1}{x^2 + 3} dx = \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
Combining these integrals, we get:
[tex]\[ \int \frac{x + 1}{x^2 + 3} dx = \frac{1}{2} \ln|x^2 + 3| + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
Thus, the integral becomes:
[tex]\[ \int \left( \frac{2}{x} - \frac{x + 1}{x^2 + 3} \right) dx \][/tex]
[tex]\[ = 2 \ln |x| - \left( \frac{1}{2} \ln |x^2 + 3| + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \right) \][/tex]
[tex]\[ = 2 \ln |x| - \frac{1}{2} \ln |x^2 + 3| - \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
### Final Answer
Thus, the partial fraction decomposition of:
[tex]\[ \frac{x^2 - x + 6}{x^3 + 3x} \][/tex]
is:
[tex]\[ \frac{2}{x} - \frac{x + 1}{x^2 + 3} \][/tex]
And the integral is:
[tex]\[ \int \frac{x^2 - x + 6}{x^3 + 3x} dx = 2 \ln |x| - \frac{1}{2} \ln |x^2 + 3| - \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
### Step 1: Partial Fraction Decomposition
We need to decompose the integrand into partial fractions.
The integrand is:
[tex]\[ \frac{x^2 - x + 6}{x^3 + 3x} \][/tex]
First, we factor the denominator:
[tex]\[ x^3 + 3x = x(x^2 + 3) \][/tex]
So, we can express the integrand as:
[tex]\[ \frac{x^2 - x + 6}{x(x^2 + 3)} \][/tex]
We assume a partial fraction decomposition of the form:
[tex]\[ \frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3} \][/tex]
Next, we multiply both sides by the denominator [tex]\( x(x^2 + 3) \)[/tex] to clear the fractions:
[tex]\[ x^2 - x + 6 = A(x^2 + 3) + (Bx + C)x \][/tex]
Now we expand and combine like terms:
[tex]\[ x^2 - x + 6 = Ax^2 + 3A + Bx^2 + Cx \][/tex]
[tex]\[ x^2 - x + 6 = (A + B)x^2 + Cx + 3A \][/tex]
We equate the coefficients of the corresponding powers of [tex]\(x\)[/tex] on both sides:
1. For [tex]\(x^2\)[/tex]: [tex]\( A + B = 1 \)[/tex]
2. For [tex]\(x\)[/tex]: [tex]\( C = -1 \)[/tex]
3. For the constant term: [tex]\( 3A = 6 \)[/tex]
From equation (3):
[tex]\[ 3A = 6 \Rightarrow A = 2 \][/tex]
Substitute [tex]\( A = 2 \)[/tex] into equation (1):
[tex]\[ 2 + B = 1 \Rightarrow B = -1 \][/tex]
So, the partial fraction decomposition is:
[tex]\[ \frac{2}{x} + \frac{(-1)x - 1}{x^2 + 3} \][/tex]
[tex]\[ = \frac{2}{x} - \frac{x + 1}{x^2 + 3} \][/tex]
### Step 2: Integrate the Partial Fractions
Now, we integrate each term separately:
[tex]\[ \int \left( \frac{2}{x} - \frac{x + 1}{x^2 + 3} \right) dx \][/tex]
First, integrate [tex]\( \frac{2}{x} \)[/tex]:
[tex]\[ \int \frac{2}{x} dx = 2 \ln |x| \][/tex]
Next, integrate [tex]\( \frac{x + 1}{x^2 + 3} \)[/tex]. Break it into two integrals:
[tex]\[ \int \frac{x + 1}{x^2 + 3} dx = \int \frac{x}{x^2 + 3} dx + \int \frac{1}{x^2 + 3} dx \][/tex]
For [tex]\( \int \frac{x}{x^2 + 3} dx \)[/tex]:
Use the substitution [tex]\( u = x^2 + 3 \)[/tex], then [tex]\( du = 2x \, dx \)[/tex]:
[tex]\[ \int \frac{x}{x^2 + 3} dx = \frac{1}{2} \int \frac{2x}{x^2 + 3} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|x^2 + 3| \][/tex]
For [tex]\( \int \frac{1}{x^2 + 3} dx \)[/tex]:
Recall the integral of [tex]\( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) \)[/tex]:
[tex]\[ \int \frac{1}{x^2 + 3} dx = \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
Combining these integrals, we get:
[tex]\[ \int \frac{x + 1}{x^2 + 3} dx = \frac{1}{2} \ln|x^2 + 3| + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
Thus, the integral becomes:
[tex]\[ \int \left( \frac{2}{x} - \frac{x + 1}{x^2 + 3} \right) dx \][/tex]
[tex]\[ = 2 \ln |x| - \left( \frac{1}{2} \ln |x^2 + 3| + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \right) \][/tex]
[tex]\[ = 2 \ln |x| - \frac{1}{2} \ln |x^2 + 3| - \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) \][/tex]
### Final Answer
Thus, the partial fraction decomposition of:
[tex]\[ \frac{x^2 - x + 6}{x^3 + 3x} \][/tex]
is:
[tex]\[ \frac{2}{x} - \frac{x + 1}{x^2 + 3} \][/tex]
And the integral is:
[tex]\[ \int \frac{x^2 - x + 6}{x^3 + 3x} dx = 2 \ln |x| - \frac{1}{2} \ln |x^2 + 3| - \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
