Question

Evaluate \( \displaystyle \int \frac{x+2}{2x^2 + 6x + 5} \, dx \) using the substitution or partial fraction method.

Hint:

If numerator is close to derivative of denominator, split it into derivative form + remainder to simplify integration.

Answer 1

Concept: When the numerator resembles the derivative of the denominator, use substitution: \[ \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C \] If not exact, split the numerator appropriately.
Step 1: Let \[ I = \int \frac{x+2}{2x^2 + 6x + 5} dx \] Differentiate denominator: \[ \frac{d}{dx}(2x^2 + 6x + 5) = 4x + 6 \] We express numerator in terms of \(4x+6\): \[ x+2 = \frac{1}{4}(4x+6) + \frac{1}{2} \]
Step 2: Split the integral \[ I = \int \frac{\frac{1}{4}(4x+6)}{2x^2+6x+5} dx + \int \frac{\frac{1}{2}}{2x^2+6x+5} dx \] \[ I = \frac{1}{4} \int \frac{4x+6}{2x^2+6x+5} dx + \frac{1}{2} \int \frac{dx}{2x^2+6x+5} \]
Step 3: First integral (log form) \[ = \frac{1}{4} \ln|2x^2+6x+5| \]
Step 4: Second integral (complete square) \[ 2x^2+6x+5 = 2(x^2+3x) + 5 \] \[ = 2\left[(x+\tfrac{3}{2})^2 + \tfrac{1}{4}\right] \] So, \[ \int \frac{dx}{2x^2+6x+5} = \frac{1}{2} \int \frac{dx}{(x+\tfrac{3}{2})^2 + (\tfrac{1}{2})^2} \] Using: \[ \int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \] \[ = \frac{1}{2} \cdot 2 \tan^{-1}(2x+3) = \tan^{-1}(2x+3) \]

Step 5: Multiply constant \[ \frac{1}{2} \times \tan^{-1}(2x+3) = \frac{1}{2}\tan^{-1}(2x+3) \] Final Answer: \[ \boxed{\frac{1}{4}\ln|2x^2+6x+5| + \frac{1}{2}\tan^{-1}(2x+3) + C} \] Explanation: The numerator was decomposed into a derivative form plus a constant. One part gave a logarithm, while the remaining quadratic produced an inverse tangent after completing the square.