t^2-3/(t-1)(t^2+1)  solove as partial fraction

To solve the expression \(\frac{t^2 - 3}{(t - 1)(t^2 + 1)}\) as a partial fraction, we decompose it into the sum of simpler fractions.

### Step-by-Step Decomposition:

1. **Given expression**:
   \[
   \frac{t^2 - 3}{(t - 1)(t^2 + 1)}
   \]

2. **Set up the partial fraction form**:
   Since the denominator is a product of \((t - 1)\) and \((t^2 + 1)\), we express the fraction as a sum of two terms:

   \[
   \frac{t^2 - 3}{(t - 1)(t^2 + 1)} = \frac{A}{t - 1} + \frac{Bt + C}{t^2 + 1}
   \]

   Where \(A\), \(B\), and \(C\) are constants to be determined.

3. **Multiply both sides by the denominator** \((t - 1)(t^2 + 1)\) to eliminate the denominators:
   \[
   t^2 - 3 = A(t^2 + 1) + (Bt + C)(t - 1)
   \]

4. **Expand the right-hand side**:
   \[
   A(t^2 + 1) = A t^2 + A
   \]
   \[
   (Bt + C)(t - 1) = Bt(t - 1) + C(t - 1) = Bt^2 - Bt + Ct - C
   \]

   So, the right-hand side becomes:
   \[
   A t^2 + A + Bt^2 - Bt + Ct - C = (A + B)t^2 + (-B + C)t + (A - C)
   \]

5. **Compare coefficients** with the left-hand side \(t^2 - 3\):
   \[
   t^2 - 3 = (A + B)t^2 + (-B + C)t + (A - C)
   \]

   Equating the coefficients of \(t^2\), \(t\), and the constant terms:

   - Coefficient of \(t^2\): \(A + B = 1\)
   - Coefficient of \(t\): \(-B + C = 0\)
   - Constant term: \(A - C = -3\)

6. **Solve the system of equations**:

   - \(A + B = 1\)
   - \(-B + C = 0\) ⟹ \(C = B\)
   - \(A - C = -3\)

   Substituting \(C = B\) into the third equation:
   \[
   A - B = -3
   \]

   Now, solve the system:
   - \(A + B = 1\)
   - \(A - B = -3\)

   Adding these two equations:
   \[
   2A = -2 \quad \Rightarrow \quad A = -1
   \]
   Substituting \(A = -1\) into \(A + B = 1\):
   \[
   -1 + B = 1 \quad \Rightarrow \quad B = 2
   \]
   Since \(C = B\), we have \(C = 2\).

7. **Final partial fraction decomposition**:
   Substituting the values of \(A\), \(B\), and \(C\) into the partial fraction form:
   \[
   \frac{t^2 - 3}{(t - 1)(t^2 + 1)} = \frac{-1}{t - 1} + \frac{2t + 2}{t^2 + 1}
   \]

Thus, the partial fraction decomposition of \(\frac{t^2 - 3}{(t - 1)(t^2 + 1)}\) is:
\[
\frac{-1}{t - 1} + \frac{2t + 2}{t^2 + 1}
\]

Would you like further details on any step, or have any specific questions?

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**Related Questions:**
1. How do we handle cases where the denominator has repeated factors in partial fractions?
2. What is the approach to decompose irreducible quadratic factors?
3. Can partial fractions be used for integrating rational functions?
4. How does the degree of the numerator affect partial fraction decomposition?
5. How would you solve this problem if the numerator had a higher degree than the denominator?

**Tip**: When decomposing into partial fractions, always ensure the degree of the numerator is less than that of the denominator; if not, perform polynomial division first.

t^2-3/(t-1)(t^2+1)  solve as partial fraction

Learn how to decompose the rational expression t^2 - 3 / (t - 1)(t^2 + 1) into partial fractions. This step-by-step solution covers partial fraction decomposition, solving systems of equations, and working with quadratic and linear factors. Suitable for advanced high school algebra students (Grades 11-12).

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