Question

Find: \[ I = \int \frac{2x^2 + 1}{x^2(x^2 + 4)} \, dx \]

Answer 1

Step 1: Use partial fractions:
\[ \frac{2x^2 + 1}{x^2(x^2 + 4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 4} \] Multiply both sides by the denominator:
\[ 2x^2 + 1 = A x (x^2 + 4) + B (x^2 + 4) + (Cx + D) x^2 \] Step 2: Expand the RHS:
\[ = Ax^3 + 4Ax + Bx^2 + 4B + Cx^3 + Dx^2 \] Group like terms: \[ = (A + C)x^3 + (B + D)x^2 + 4Ax + 4B \] Step 3: Compare coefficients:

• Coefficient of \( x^3 \): \( A + C = 0 \)
• Coefficient of \( x^2 \): \( B + D = 2 \)
• Coefficient of \( x \): \( 4A = 0 \Rightarrow A = 0 \)
• Constant term: \( 4B = 1 \Rightarrow B = \frac{1}{4} \)

From above: \[ A = 0 \Rightarrow C = 0,\quad B = \frac{1}{4} \Rightarrow D = 2 - \frac{1}{4} = \frac{7}{4} \] Step 4: So the decomposition becomes: \[ \frac{2x^2 + 1}{x^2(x^2 + 4)} = \frac{1}{4x^2} + \frac{7}{4(x^2 + 4)} \] Step 5: Now integrate: \[ I = \int \left( \frac{1}{4x^2} + \frac{7}{4(x^2 + 4)} \right) dx \] Step 6: Use standard integrals: \[ \int \frac{1}{x^2} dx = -\frac{1}{x},\quad \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \] So: \[ I = -\frac{1}{4x} + \frac{7}{4} \cdot \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C = -\frac{1}{4x} + \frac{7}{8} \tan^{-1} \left( \frac{x}{2} \right) + C \] Final Answer:
\[ \boxed{I = -\frac{1}{4x} + \frac{7}{8} \tan^{-1} \left( \frac{x}{2} \right) + C} \]