Question

Find the value of the integral: \[ I = \int \frac{x^2 + x + 1}{(x+2)(x^2 + 1)} dx. \]

Hint:

For integrating rational functions, always attempt partial fraction decomposition. Look for linear and quadratic factors to decompose and simplify.

Answer 1

We use partial fraction decomposition to split the given 6266899d2bbfcb1799af2d57. Assume: \[ \frac{x^2 + x + 1}{(x+2)(x^2 + 1)} = \frac{A}{x+2} + \frac{Bx + C}{x^2 + 1}. \] Multiplying both sides by \((x+2)(x^2 + 1)\), we get: \[ x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x+2). \] Expanding the right-hand side: \[ x^2 + x + 1 = A x^2 + A + Bx^2 + 2Bx + Cx + 2C. \] \[ = (A + B)x^2 + (2B + C)x + (A + 2C). \] Equating coefficients, we get: 1. \( A + B = 1 \)
2. \( 2B + C = 1 \)
3. \( A + 2C = 1 \) Solving these equations: - From (1), \( A = 1 - B \).
- Substituting into (3): \( (1 - B) + 2C = 1 \Rightarrow 2C = B \Rightarrow C = \frac{B}{2} \).
- Substituting into (2): \( 2B + \frac{B}{2} = 1 \Rightarrow \frac{4B + B}{2} = 1 \Rightarrow 5B = 2 \Rightarrow B = \frac{2}{5} \).
- So, \( A = 1 - \frac{2}{5} = \frac{3}{5} \).
- And, \( C = \frac{1}{5} \). Thus, we rewrite: \[ I = \int \left( \frac{3/5}{x+2} + \frac{(2/5)x + (1/5)}{x^2 + 1} \right) dx. \] \[ = \frac{3}{5} \int \frac{dx}{x+2} + \frac{2}{5} \int \frac{x dx}{x^2 + 1} + \frac{1}{5} \int \frac{dx}{x^2 + 1}. \] Solving these integrals: \[ I = \frac{3}{5} \ln |x+2| + \frac{2}{5} \cdot \frac{1}{2} \ln |x^2 + 1| + \frac{1}{5} \tan^{-1} x + C. \] \[ = \frac{3}{5} \ln |x+2| + \frac{1}{5} \ln |x^2 + 1| + \frac{1}{5} \tan^{-1} x + C. \]