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Sagot :
Certainly! To perform partial fraction decomposition on the expression [tex]\(\frac{1 - 2x}{x^2 + 1}\)[/tex], we need to express it in a form where it can be separated into simpler rational functions. Here are the detailed steps:
1. Identify the structure:
The given expression is [tex]\(\frac{1 - 2x}{x^2 + 1}\)[/tex].
2. Check the denominator:
The denominator [tex]\(x^2 + 1\)[/tex] is a quadratic polynomial. It does not factorize into simpler linear terms over the real numbers, but remains as [tex]\(x^2 + 1\)[/tex].
3. Set up the partial fraction decomposition:
Since [tex]\(x^2 + 1\)[/tex] is an irreducible quadratic expression, we propose that our partial fraction decomposition will take the form:
[tex]\[ \frac{1 - 2x}{x^2 + 1} = \frac{Ax + B}{x^2 + 1} \][/tex]
4. Equate the numerators:
Multiply both sides by [tex]\(x^2 + 1\)[/tex] to clear the denominator:
[tex]\[ 1 - 2x = Ax + B \][/tex]
5. Solve for variables [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
By equating the coefficients of corresponding powers of [tex]\(x\)[/tex] on both sides of the equation, we get:
- For the constant term: [tex]\(1 = B\)[/tex]
- For the coefficient of [tex]\(x\)[/tex]: [tex]\(-2 = A\)[/tex]
Thus, we find that [tex]\(A = -2\)[/tex] and [tex]\(B = 1\)[/tex].
6. Re-write the partial fraction decomposition:
Plug [tex]\(A\)[/tex] and [tex]\(B\)[/tex] back into the proposed partial fraction decomposition form:
[tex]\[ \frac{1 - 2x}{x^2 + 1} = \frac{-2x + 1}{x^2 + 1} \][/tex]
7. Final result:
Notice that the partial fraction decomposition results in the same form as the original expression because the coefficients correctly represent the given structure.
Thus, the partial fraction decomposition of [tex]\(\frac{1 - 2x}{x^2 + 1}\)[/tex] is:
[tex]\[ \frac{1 - 2x}{x^2 + 1} = -\frac{2x - 1}{x^2 + 1} \][/tex]
1. Identify the structure:
The given expression is [tex]\(\frac{1 - 2x}{x^2 + 1}\)[/tex].
2. Check the denominator:
The denominator [tex]\(x^2 + 1\)[/tex] is a quadratic polynomial. It does not factorize into simpler linear terms over the real numbers, but remains as [tex]\(x^2 + 1\)[/tex].
3. Set up the partial fraction decomposition:
Since [tex]\(x^2 + 1\)[/tex] is an irreducible quadratic expression, we propose that our partial fraction decomposition will take the form:
[tex]\[ \frac{1 - 2x}{x^2 + 1} = \frac{Ax + B}{x^2 + 1} \][/tex]
4. Equate the numerators:
Multiply both sides by [tex]\(x^2 + 1\)[/tex] to clear the denominator:
[tex]\[ 1 - 2x = Ax + B \][/tex]
5. Solve for variables [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
By equating the coefficients of corresponding powers of [tex]\(x\)[/tex] on both sides of the equation, we get:
- For the constant term: [tex]\(1 = B\)[/tex]
- For the coefficient of [tex]\(x\)[/tex]: [tex]\(-2 = A\)[/tex]
Thus, we find that [tex]\(A = -2\)[/tex] and [tex]\(B = 1\)[/tex].
6. Re-write the partial fraction decomposition:
Plug [tex]\(A\)[/tex] and [tex]\(B\)[/tex] back into the proposed partial fraction decomposition form:
[tex]\[ \frac{1 - 2x}{x^2 + 1} = \frac{-2x + 1}{x^2 + 1} \][/tex]
7. Final result:
Notice that the partial fraction decomposition results in the same form as the original expression because the coefficients correctly represent the given structure.
Thus, the partial fraction decomposition of [tex]\(\frac{1 - 2x}{x^2 + 1}\)[/tex] is:
[tex]\[ \frac{1 - 2x}{x^2 + 1} = -\frac{2x - 1}{x^2 + 1} \][/tex]
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