Question

Evaluate the integral: \[ \int \frac{x}{x+2}\, dx \]

• A. \(x - 2\ln|x+2| + C \)
• B. \(x + 2\ln|x+2| + C \)
• C. \(x - \ln|x+2| + C \)
• D. \( \ln|x+2| + C \)

Hint:

For integrals of the form \( \frac{x}{x+a} \), rewrite the numerator as \(x+a-a\). This converts the expression into \(1 - \frac{a}{x+a}\), making the integration straightforward.

Answer 1

The Correct Option is A

Concept: When the degree of the numerator is equal to or greater than the denominator, we simplify the integrand using algebraic division or decomposition. A useful trick is to rewrite the numerator so that the denominator appears in the expression. \[ \frac{x}{x+2} \] can be rewritten as \[ \frac{x+2-2}{x+2} \] which simplifies the integration process.
Step 1: {Rewrite the numerator.} \[ \frac{x}{x+2} = \frac{x+2-2}{x+2} \] \[ = \frac{x+2}{x+2} - \frac{2}{x+2} \] \[ = 1 - \frac{2}{x+2} \] Thus the integral becomes \[ \int \frac{x}{x+2}dx = \int \left(1 - \frac{2}{x+2}\right)dx \]
Step 2: {Integrate term by term.} \[ \int 1\,dx = x \] \[ \int \frac{1}{x+2}dx = \ln|x+2| \] Therefore, \[ \int \left(1 - \frac{2}{x+2}\right)dx \] \[ = x - 2\ln|x+2| + C \]
Step 3: {Write the final result.} \[ \int \frac{x}{x+2}\,dx = x - 2\ln|x+2| + C \]