Question

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

Hint:

To integrate rational functions, use partial fraction decomposition to split the expression into simpler terms.

Answer 1

Step 1: Use partial fraction decomposition to simplify the integrand: \[ \frac{2x^2 - 3}{(x^2 - 5)(x^2 + 4)} = \frac{Ax + B}{x^2 - 5} + \frac{Cx + D}{x^2 + 4} \] Step 2: Multiply both sides by \( (x^2 - 5)(x^2 + 4) \) to eliminate the denominators: \[ 2x^2 - 3 = (Ax + B)(x^2 + 4) + (Cx + D)(x^2 - 5) \] Step 3: Expand both sides: \[ 2x^2 - 3 = (Ax^3 + 4Ax + Bx^2 + 4B) + (Cx^3 - 5Cx + Dx^2 - 5D) \] \[ = (A + C)x^3 + (B + D)x^2 + (4A - 5C)x + (4B - 5D) \] Step 4: Set up a system of equations by equating the coefficients of like powers of \( x \): - Coefficient of \( x^3 \): \( A + C = 0 \) - Coefficient of \( x^2 \): \( B + D = 2 \) - Coefficient of \( x \): \( 4A - 5C = 0 \) - Constant term: \( 4B - 5D = -3 \) Step 5: Solve the system of equations: From \( A + C = 0 \), we get \( C = -A \). Substitute into \( 4A - 5C = 0 \): \[ 4A - 5(-A) = 0 \quad \Rightarrow \quad 9A = 0 \quad \Rightarrow \quad A = 0, \quad C = 0 \] From \( B + D = 2 \), we get \( D = 2 - B \). Substitute into \( 4B - 5D = -3 \): \[ 4B - 5(2 - B) = -3 \quad \Rightarrow \quad 4B - 10 + 5B = -3 \] \[ 9B = 7 \quad \Rightarrow \quad B = \frac{7}{9}, \quad D = 2 - \frac{7}{9} = \frac{11}{9} \] Step 6: The partial fractions are: \[ \frac{2x^2 - 3}{(x^2 - 5)(x^2 + 4)} = \frac{7}{9(x^2 - 5)} + \frac{11}{9(x^2 + 4)} \] Step 7: Now, integrate term by term: \[ \int \frac{7}{9(x^2 - 5)} \, dx + \int \frac{11}{9(x^2 + 4)} \, dx \] The integrals are standard: \[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| \] Thus, the result is: \[ \frac{7}{9} \cdot \frac{1}{2\sqrt{5}} \ln \left| \frac{x - \sqrt{5}}{x + \sqrt{5}} \right| + \frac{11}{9} \cdot \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C \]