Question

$\int \frac{x^2 + 6x + 1}{(x+3)^2} dx = $}

• A. $x - \frac{8}{x+3} + C$
• B. $x - \frac{16}{x+3} + C$
• C. $x + \frac{8}{x+3} + C$
• D. $x + \frac{16}{x+3} + C$
• E. $x + \frac{4}{x+3} + C$

Hint:

For rational integrals where the denominator is a perfect square, look at the coefficients of $x^2$ and $x$. If they match the denominator's expansion, simply adjust the constant term to split the integral.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are integrating a rational function where the degree of the numerator equals the degree of the denominator.
We should perform polynomial division or algebra manipulation to separate it into a polynomial and a simpler fraction.
Step 2: Key Formula or Approach:
Notice the denominator is \((x + 3)^2 = x^2 + 6x + 9\).
Add and subtract 8 in the numerator to match the expanded denominator.
Integrate the separated terms.
Step 3: Detailed Explanation:
Expand the denominator:
\[ (x + 3)^2 = x^2 + 6x + 9 \] Rewrite the numerator to include this term:
\[ x^2 + 6x + 1 = (x^2 + 6x + 9) - 8 \] \[ = (x + 3)^2 - 8 \] Now rewrite the integral:
\[ I = \int \frac{(x + 3)^2 - 8}{(x + 3)^2} dx \] Split the fraction:
\[ I = \int \left( \frac{(x + 3)^2}{(x + 3)^2} - \frac{8}{(x + 3)^2} \right) dx \] \[ I = \int \left( 1 - 8(x + 3)^{-2} \right) dx \] Integrate each part:
\[ I = x - 8 \frac{(x + 3)^{-1}}{-1} + C \] \[ I = x + \frac{8}{x + 3} + C \] Step 4: Final Answer:
The result is \(x + \frac{8}{x + 3} + C\).