Answer :
To simplify the expression [tex]\(\frac{x^2 + 1}{x^2 - 1}\)[/tex] using partial fractions, follow these steps:
1. Factor the Denominator:
The denominator [tex]\(x^2 - 1\)[/tex] can be factored as [tex]\((x - 1)(x + 1)\)[/tex]. So, the given expression becomes:
[tex]\[ \frac{x^2 + 1}{(x - 1)(x + 1)} \][/tex]
2. Set Up the Partial Fraction Decomposition:
We want to express the fraction as a sum of simpler fractions. Assume the decomposition to be:
[tex]\[ \frac{x^2 + 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants to be determined.
3. Combine the Right-Hand Side into a Single Fraction:
[tex]\[ \frac{A}{x - 1} + \frac{B}{x + 1} = \frac{A(x + 1) + B(x - 1)}{(x - 1)(x + 1)} \][/tex]
Simplify the numerator on the right-hand side:
[tex]\[ A(x + 1) + B(x - 1) = Ax + A + Bx - B = (A + B)x + (A - B) \][/tex]
4. Equate the Numerators:
Now we match the numerator of the original fraction with the numerator of the combined fraction:
[tex]\[ x^2 + 1 = (A + B)x + (A - B) \][/tex]
For these to be equal for all [tex]\(x\)[/tex], the coefficients of corresponding powers of [tex]\(x\)[/tex] on both sides must be equal:
For [tex]\(x\)[/tex]:
[tex]\[ 0 = A + B \][/tex]
For the constant term:
[tex]\[ 1 = A - B \][/tex]
5. Solve the System of Equations:
Solve the system of equations to find [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
From [tex]\(0 = A + B\)[/tex]:
[tex]\[ B = -A \][/tex]
Substitute [tex]\(B = -A\)[/tex] into [tex]\(1 = A - B\)[/tex]:
[tex]\[ 1 = A - (-A) \][/tex]
[tex]\[ 1 = 2A \][/tex]
[tex]\[ A = \frac{1}{2} \][/tex]
Substituting [tex]\(A = \frac{1}{2}\)[/tex] back into [tex]\(B = -A\)[/tex]:
[tex]\[ B = -\frac{1}{2} \][/tex]
6. Write the Partial Fraction Decomposition:
Substitute [tex]\(A\)[/tex] and [tex]\(B\)[/tex] back into the partial fractions form:
[tex]\[ \frac{x^2 + 1}{(x - 1)(x + 1)} = \frac{\frac{1}{2}}{x - 1} + \frac{-\frac{1}{2}}{x + 1} \][/tex]
Simplify the fractions:
[tex]\[ \frac{1}{2(x - 1)} - \frac{1}{2(x + 1)} \][/tex]
This can be further simplified, considering the properties of fractions:
[tex]\[ \left(\frac{1}{x-1} - \frac{1}{x+1} \right) / 2 \][/tex]
So, the partial fraction decomposition of [tex]\(\frac{x^2 + 1}{x^2 - 1}\)[/tex] is:
[tex]\[ \frac{1}{x - 1} - \frac{1}{x + 1} \][/tex]
Putting it all together:
[tex]\[ \boxed{1 - \frac{1}{x + 1} + \frac{1}{x - 1}} \][/tex]
1. Factor the Denominator:
The denominator [tex]\(x^2 - 1\)[/tex] can be factored as [tex]\((x - 1)(x + 1)\)[/tex]. So, the given expression becomes:
[tex]\[ \frac{x^2 + 1}{(x - 1)(x + 1)} \][/tex]
2. Set Up the Partial Fraction Decomposition:
We want to express the fraction as a sum of simpler fractions. Assume the decomposition to be:
[tex]\[ \frac{x^2 + 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants to be determined.
3. Combine the Right-Hand Side into a Single Fraction:
[tex]\[ \frac{A}{x - 1} + \frac{B}{x + 1} = \frac{A(x + 1) + B(x - 1)}{(x - 1)(x + 1)} \][/tex]
Simplify the numerator on the right-hand side:
[tex]\[ A(x + 1) + B(x - 1) = Ax + A + Bx - B = (A + B)x + (A - B) \][/tex]
4. Equate the Numerators:
Now we match the numerator of the original fraction with the numerator of the combined fraction:
[tex]\[ x^2 + 1 = (A + B)x + (A - B) \][/tex]
For these to be equal for all [tex]\(x\)[/tex], the coefficients of corresponding powers of [tex]\(x\)[/tex] on both sides must be equal:
For [tex]\(x\)[/tex]:
[tex]\[ 0 = A + B \][/tex]
For the constant term:
[tex]\[ 1 = A - B \][/tex]
5. Solve the System of Equations:
Solve the system of equations to find [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
From [tex]\(0 = A + B\)[/tex]:
[tex]\[ B = -A \][/tex]
Substitute [tex]\(B = -A\)[/tex] into [tex]\(1 = A - B\)[/tex]:
[tex]\[ 1 = A - (-A) \][/tex]
[tex]\[ 1 = 2A \][/tex]
[tex]\[ A = \frac{1}{2} \][/tex]
Substituting [tex]\(A = \frac{1}{2}\)[/tex] back into [tex]\(B = -A\)[/tex]:
[tex]\[ B = -\frac{1}{2} \][/tex]
6. Write the Partial Fraction Decomposition:
Substitute [tex]\(A\)[/tex] and [tex]\(B\)[/tex] back into the partial fractions form:
[tex]\[ \frac{x^2 + 1}{(x - 1)(x + 1)} = \frac{\frac{1}{2}}{x - 1} + \frac{-\frac{1}{2}}{x + 1} \][/tex]
Simplify the fractions:
[tex]\[ \frac{1}{2(x - 1)} - \frac{1}{2(x + 1)} \][/tex]
This can be further simplified, considering the properties of fractions:
[tex]\[ \left(\frac{1}{x-1} - \frac{1}{x+1} \right) / 2 \][/tex]
So, the partial fraction decomposition of [tex]\(\frac{x^2 + 1}{x^2 - 1}\)[/tex] is:
[tex]\[ \frac{1}{x - 1} - \frac{1}{x + 1} \][/tex]
Putting it all together:
[tex]\[ \boxed{1 - \frac{1}{x + 1} + \frac{1}{x - 1}} \][/tex]