Question

Find the indefinite integral: \[ \int \frac{2x + 1}{x^2 + 6x + 2} \, dx \]

Hint:

Always double-check your completed square form by re-expanding it: \((x+3)^2 - 7 = x^2 + 6x + 9 - 7 = x^2 + 6x + 2\). This guarantees that no sign errors were introduced midway.

Answer 1

Concept: The given integral is of the standard rational form: \[ \int \frac{px + q}{ax^2 + bx + c} \, dx \] To solve this, we express the linear numerator expression as a linear combination involving the derivative of the quadratic denominator expression plus a constant offset term. That is: \[ \text{Numerator} = A \cdot \frac{d}{dx}(\text{Denominator}) + B \] Once the constants \(A\) and \(B\) are found, the expression breaks down cleanly into two separate integrals:
• \(\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C\)
• An integral of the form \(\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C\), evaluated by completing the square in the quadratic expression.

Step 1: Expressing the numerator in terms of the derivative of the denominator.
Let the given integral be \(I\): \[ I = \int \frac{2x + 1}{x^2 + 6x + 2} \, dx \] The derivative of the denominator quadratic expression with respect to \(x\) is: \[ \frac{d}{dx}(x^2 + 6x + 2) = 2x + 6 \] We now establish the identity for the numerator: \[ 2x + 1 = A(2x + 6) + B \] Expanding the right-hand side of the identity: \[ 2x + 1 = 2Ax + (6A + B) \] By equating the corresponding coefficients of \(x\) on both sides: \[ 2A = 2 \quad \Rightarrow \quad A = 1 \] By equating the constant terms on both sides: \[ 6A + B = 1 \] Substitute \(A = 1\) into this constant relation: \[ 6(1) + B = 1 \quad \Rightarrow \quad B = 1 - 6 = -5 \] Therefore, we can rewrite the numerator explicitly as: \[ 2x + 1 = 1(2x + 6) - 5 \]

Step 2: Splitting the primary integral into two sub-integrals.
Substitute the newly constructed numerator back into the integral expression \(I\): \[ I = \int \frac{(2x + 6) - 5}{x^2 + 6x + 2} \, dx \] Splitting this up across the numerator subtraction yields: \[ I = \int \frac{2x + 6}{x^2 + 6x + 2} \, dx - 5 \int \frac{1}{x^2 + 6x + 2} \, dx \] Let us denote these two sub-integrals as \(I_1\) and \(I_2\) respectively, so that: \[ I = I_1 - 5I_2 \quad \cdots (1) \]

Step 3: Evaluating the first sub-integral \(I_1\).
\[ I_1 = \int \frac{2x + 6}{x^2 + 6x + 2} \, dx \] Let us substitute \(t = x^2 + 6x + 2\). Taking differentials on both sides: \[ dt = (2x + 6) \, dx \] Substituting these values converts the integral to: \[ I_1 = \int \frac{1}{t} \, dt = \ln|t| = \ln|x^2 + 6x + 2| + C_1 \quad \cdots (2) \]

Step 4: Evaluating the second sub-integral \(I_2\) by completing the square.
\[ I_2 = \int \frac{1}{x^2 + 6x + 2} \, dx \] To complete the square for the quadratic expression in the denominator, look at the coefficient of \(x\), which is 6. Halving it gives 3, and squaring that gives 9. Thus, we add and subtract 9 inside the expression: \[ x^2 + 6x + 2 = (x^2 + 6x + 9) - 9 + 2 = (x+3)^2 - 7 \] We can express 7 as an exact square term: \(7 = (\sqrt{7})^2\). Thus: \[ I_2 = \int \frac{1}{(x+3)^2 - (\sqrt{7})^2} \, dx \] Now, apply the standard standard integral formula \(\int \frac{1}{u^2 - a^2} \, du = \frac{1}{2a} \ln\left|\frac{u-a}{u+a}\right|\), where \(u = x+3\) and \(a = \sqrt{7}\): \[ I_2 = \frac{1}{2\sqrt{7}} \ln\left| \frac{(x+3) - \sqrt{7}}{(x+3) + \sqrt{7}} \right| + C_2 \quad \cdots (3) \]

Step 5: Consolidating the parts to get the final integration output.
Substitute equations (2) and (3) back into equation (1), and combine the integration constants \(C_1 - 5C_2\) into a single final constant \(C\): \[ I = \ln|x^2 + 6x + 2| - 5 \left( \frac{1}{2\sqrt{7}} \ln\left| \frac{x+3 - \sqrt{7}}{x+3 + \sqrt{7}} \right| \right) + C \] \[ I = \ln|x^2 + 6x + 2| - \frac{5}{2\sqrt{7}} \ln\left| \frac{x+3 - \sqrt{7}}{x+3 + \sqrt{7}} \right| + C \] This fully matches option (A).