Question

Evaluate \( \int \frac{x^2 + 6x + 1}{(x+3)^2} \, dx \).

Hint:

When encountering rational functions, simplify the expression before integrating, and use standard integral formulas like \( \int \frac{1}{x^2} \, dx = -\frac{1}{x} \).

Answer 1

Step 1: Simplify the integrand.
We start by simplifying the numerator. We can divide \( x^2 + 6x + 1 \) by \( (x+3)^2 \) using polynomial division. Performing the division: \[ \frac{x^2 + 6x + 1}{(x+3)^2} = \frac{(x+3)^2 + 4}{(x+3)^2} = 1 + \frac{4}{(x+3)^2} \]
Step 2: Break the integral into simpler parts.
The integral becomes: \[ \int \left( 1 + \frac{4}{(x+3)^2} \right) dx \]
Step 3: Integrate each term.
First, integrate \( 1 \): \[ \int 1 \, dx = x \] Next, integrate \( \frac{4}{(x+3)^2} \): \[ \int \frac{4}{(x+3)^2} \, dx = -\frac{4}{x+3} \]
Step 4: Combine the results.
The integral is: \[ x - \frac{4}{x+3} + C \]