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 can simplify the numerator to involve the factor in the denominator by completing the square or by long division.

Step 2: Detailed Explanation:
Note that the square of the binomial in the denominator is \( (x+3)^2 = x^2 + 6x + 9 \).
We can rewrite the numerator as:
\[ x^2 + 6x + 1 = (x^2 + 6x + 9) - 8 = (x+3)^2 - 8 \]
Substitute this into the integral:
\[ I = \int \frac{(x+3)^2 - 8}{(x+3)^2} dx = \int \left( 1 - \frac{8}{(x+3)^2} \right) dx \]
Integrate term by term:
\[ I = \int 1 dx - 8 \int (x+3)^{-2} dx \]
\[ I = x - 8 \left( \frac{(x+3)^{-1}}{-1} \right) + C \]
\[ I = x + \frac{8}{x+3} + C \]

Step 3: Final Answer:
The integral is \( x + \frac{8}{x+3} + C \).