3. \( \int \frac{\left(3 x^{3}-4 x^{2}+3 x\right)}{x^{2}+1} d x \)

To solve the integral \( \int \frac{3 x^{3} - 4 x^{2} + 3 x}{x^{2} + 1} \, dx \), we first perform polynomial long division, as the degree of the numerator is greater than the degree of the denominator. 1. Divide \( 3 x^{3} - 4 x^{2} + 3 x \) by \( x^{2}+1 \): - The first term: \( 3x \) (since \( \frac{3x^3}{x^2} = 3x \)). - Multiply \( 3x \) by \( x^2 + 1 \) yielding \( 3x^3 + 3x \). - Subtract: \( (3x^3 - 4x^2 + 3x) - (3x^3 + 3x) = -4x^2 \). 2. Now, we have \( -4x^2 \) left, which we can now integrate separately: \[ \int \left( 3x + \frac{-4x^2}{x^2 + 1} \right) dx = \int 3x \, dx + \int \frac{-4x^2}{x^2 + 1} \, dx. \] 3. The first integral is straightforward: \[ \int 3x \, dx = \frac{3x^2}{2} + C_1. \] 4. For the second part, we can simplify: \[ \frac{-4x^2}{x^2 + 1} = -4 + \frac{4}{x^2 + 1}. \] 5. Thus, we need to integrate: \[ \int -4 \, dx + \int \frac{4}{x^2 + 1} \, dx = -4x + 4 \tan^{-1}(x) + C_2. \] 6. Combine all parts: \[ \int \frac{3 x^{3} - 4 x^{2} + 3 x}{x^{2} + 1} \, dx = \frac{3x^2}{2} - 4x + 4 \tan^{-1}(x) + C. \] Therefore, the final answer is: \[ \frac{3x^2}{2} - 4x + 4 \tan^{-1}(x) + C. \]