Question

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

Hint:

Quick Tip: When solving integrals involving rational functions, consider partial fraction decomposition. It can simplify the integrand and make the integral easier to evaluate.

Answer 1

We are given the integral: \[ I = \int \frac{x^2 + 1}{(x^2 + 2)(2x^2 + 1)} \, dx \] To solve this integral, we will use partial fraction decomposition. Let's decompose the rational function: \[ \frac{x^2 + 1}{(x^2 + 2)(2x^2 + 1)} = \frac{A}{x^2 + 2} + \frac{B}{2x^2 + 1} \] Now, we need to find the values of \( A \) and \( B \). Multiply both sides by \( (x^2 + 2)(2x^2 + 1) \) to clear the denominators: \[ x^2 + 1 = A(2x^2 + 1) + B(x^2 + 2) \] Expanding both sides: \[ x^2 + 1 = A(2x^2 + 1) + B(x^2 + 2) \] \[ x^2 + 1 = A(2x^2) + A + Bx^2 + 2B \] \[ x^2 + 1 = (2A + B)x^2 + (A + 2B) \] Now, equate the coefficients of \( x^2 \) and the constant terms on both sides: 1. \( 2A + B = 1 \) 2. \( A + 2B = 1 \) Solve this system of equations: From equation (1), \( B = 1 - 2A \). Substitute into equation (2): \[ A + 2(1 - 2A) = 1 \] \[ A + 2 - 4A = 1 \] \[ -3A = -1 \quad \Rightarrow \quad A = \frac{1}{3} \] Substitute \( A = \frac{1}{3} \) into \( B = 1 - 2A \): \[ B = 1 - 2 \times \frac{1}{3} = \frac{1}{3} \] Thus, we have \( A = \frac{1}{3} \) and \( B = \frac{1}{3} \). So, the integral becomes: \[ I = \frac{1}{3} \int \frac{1}{x^2 + 2} \, dx + \frac{1}{3} \int \frac{1}{2x^2 + 1} \, dx \] Now, use standard integration formulas: \[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \] The first integral is: \[ \int \frac{dx}{x^2 + 2} = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \] The second integral is: \[ \int \frac{dx}{2x^2 + 1} = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \] Thus, the final answer is: \[ I = \frac{1}{3} \left( \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) + \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \right) + C \]