Question

Evaluate the integral: \(\int \frac{x}{x + 2} \, dx\).

Hint:

Use the denominator as the new variable: set $u=x+2$. Then replace $x$ by $u-2$ before simplifying and integrating.

Answer 1

Step 1: Understanding the Concept:
When the degree of the numerator is greater than or equal to the degree of the denominator, we use algebraic manipulation or division.

Step 2: Key Formula or Approach:
1. Addition and subtraction in the numerator: \(\frac{x}{x+2} = \frac{x+2-2}{x+2}\).
2. \(\int \frac{1}{x+a} dx = \log|x+a| + C\).

Step 3: Detailed Explanation:
Rewrite the expression inside the integral:
\[ \frac{x}{x+2} = \frac{(x+2) - 2}{x+2} = \frac{x+2}{x+2} - \frac{2}{x+2} = 1 - \frac{2}{x+2} \] Integrate:
\[ \int \left(1 - \frac{2}{x+2}\right) dx = \int 1 dx - 2 \int \frac{1}{x+2} dx \] \[ = x - 2\log|x+2| + C \]
Step 4: Final Answer:
The integral is \(x - 2\log|x + 2| + C\).

Answer 2

Concept:

• A substitution based on the denominator converts the rational expression into a simple power of the new variable.
• After setting $u=x+2$, express both $x$ and $dx$ in terms of $u$.

Step 1: Substitute for the denominator.
Let $u=x+2$. Then $x=u-2$ and $du=dx$.

Step 2: Rewrite the integral in terms of $u$.
$\int\dfrac{x}{x+2}\,dx=\int\dfrac{u-2}{u}\,du$
$=\int\left(1-\dfrac2u\right)du$

Step 3: Integrate the two terms.
$\int1\,du-2\int\dfrac1u\,du=u-2\ln|u|+C$

Step 4: Replace $u$ by $x+2$.
$u-2\ln|u|+C=x+2-2\ln|x+2|+C$
The constant $2$ is absorbed into the arbitrary integration constant.

Final Answer: $x-2\ln|x+2|+C$