Question

Evaluate: \( \int \frac{2x^2 - 3}{(x^2 - 5)(x + 4) \, dx \).
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Hint:

Use partial fractions for rational functions; split numerators for complex denominators.

Answer 1

Use partial fractions: \( \frac{2x^2 - 3}{(x^2 - 5)(x + 4)} = \frac{A}{x + 4} + \frac{Bx + C}{x^2 - 5} \).
\[ 2x^2 - 3 = A(x^2 - 5) + (Bx + C)(x + 4). \] Equate coefficients:
- \( x^2 \): \( A + B = 2 \).
- \( x \): \( 4B + C = 0 \).
- Constant: \( -5A + 4C = -3 \).
Solve: From \( 4B + C = 0 \), \( C = -4B \).
\[ -5A + 4(-4B) = -3 \Rightarrow -5A - 16B = -3. \] \[ A + B = 2 \Rightarrow A = 2 - B. \] \[ -5(2 - B) - 16B = -3 \Rightarrow -10 + 5B - 16B = -3 \Rightarrow -11B = 7 \Rightarrow B = -\frac{7}{11}. \] \[ A = 2 - \left(-\frac{7}{11}\right) = \frac{29}{11}, \quad C = -4 \cdot \left(-\frac{7}{11}\right) = \frac{28}{11}. \] \[ \int \left( \frac{\frac{29}{11}}{x + 4} + \frac{-\frac{7}{11}x + \frac{28}{11}}{x^2 - 5} \right) dx = \frac{29}{11} \int \frac{1}{x + 4} \, dx + \frac{1}{11} \int \frac{-7x + 28}{x^2 - 5} \, dx. \] \[ = \frac{29}{11} \ln |x + 4| + \frac{1}{11} \int \frac{-7x + 28}{x^2 - 5} \, dx. \] For \( \int \frac{-7x + 28}{x^2 - 5} \, dx \): \[ \int \frac{-7x}{x^2 - 5} \, dx = -\frac{7}{2} \ln (x^2 - 5), \quad \int \frac{28}{x^2 - 5} \, dx = \frac{28}{\sqrt{5}} \ln \left| \frac{x - \sqrt{5}}{x + \sqrt{5}} \right|. \] Combine: \[ \frac{29}{11} \ln |x + 4| - \frac{7}{22} \ln (x^2 - 5) + \frac{28}{11\sqrt{5}} \ln \left| \frac{x - \sqrt{5}}{x + \sqrt{5}} \right| + c. \] Answer: \( \frac{29}{11} \ln |x + 4| - \frac{7}{22} \ln (x^2 - 5) + \frac{28}{11\sqrt{5}} \ln \left| \frac{x - \sqrt{5}}{x + \sqrt{5}} \right| + c \).