\( \int \frac { x + 2 } { x ( x ^ { 2 } + x + 1 ) } d x = \int \)

To evaluate the integral: \[ \int \frac{x + 2}{x(x^2 + x + 1)} \, dx \] we can use the method of **partial fractions**. Here's a step-by-step solution: ### 1. Break Down the Fraction Express the integrand as a sum of simpler fractions: \[ \frac{x + 2}{x(x^2 + x + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + x + 1} \] ### 2. Determine Constants \( A \), \( B \), and \( C \) Multiply both sides by the denominator \( x(x^2 + x + 1) \): \[ x + 2 = A(x^2 + x + 1) + (Bx + C)x \] Expand and collect like terms: \[ x + 2 = (A + B)x^2 + (A + C)x + A \] Set up equations by equating the coefficients of corresponding powers of \( x \): \[ \begin{cases} A + B = 0 \quad &\text{(coefficient of } x^2) \\ A + C = 1 \quad &\text{(coefficient of } x) \\ A = 2 \quad &\text{(constant term)} \end{cases} \] Solve the system: \[ \begin{align*} A &= 2 \\ B &= -2 \quad (\text{since } A + B = 0) \\ C &= -1 \quad (\text{since } A + C = 1) \end{align*} \] ### 3. Rewrite the Integral Substitute \( A \), \( B \), and \( C \) back into the partial fractions: \[ \frac{x + 2}{x(x^2 + x + 1)} = \frac{2}{x} + \frac{-2x - 1}{x^2 + x + 1} \] So the integral becomes: \[ \int \left( \frac{2}{x} + \frac{-2x - 1}{x^2 + x + 1} \right) dx \] ### 4. Integrate Term by Term **First Term:** \[ \int \frac{2}{x} \, dx = 2 \ln |x| \] **Second Term:** Let \( u = x^2 + x + 1 \). Then, \[ du = (2x + 1) \, dx \quad \Rightarrow \quad -du = (-2x - 1) \, dx \] Thus, \[ \int \frac{-2x - 1}{x^2 + x + 1} \, dx = -\int \frac{du}{u} = -\ln |u| = -\ln (x^2 + x + 1) \] ### 5. Combine the Results Putting it all together: \[ \int \frac{x + 2}{x(x^2 + x + 1)} \, dx = 2 \ln |x| - \ln (x^2 + x + 1) + C \] where \( C \) is the constant of integration. ### Final Answer \[ \int \frac{x + 2}{x(x^2 + x + 1)} \, dx = 2 \ln|x| - \ln(x^2 + x + 1) + C \]