$\int\frac{1}{(x + 1)(x+2)}dx$ is equal to

$\log_e\left|\frac{x+1}{x+2}\right|+C$: where $C$ is an arbitrary constant

The correct answer is Option (2) → $\log_e\left|\frac{x+1}{x+2}\right|+C$: where $C$ is an arbitrary constant

Given integral: $\int \frac{1}{(x+1)(x+2)} dx$

Use partial fraction decomposition:

$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$

Multiply both sides by (x+1)(x+2):

1 = A(x+2) + B(x+1) = Ax + 2A + Bx + B = (A+B)x + (2A+B)

Compare coefficients:

A + B = 0 ⇒ B = -A

2A + B = 1 ⇒ 2A - A = 1 ⇒ A = 1 ⇒ B = -1

Thus:

$\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}$

Integrate:

$\int \frac{1}{(x+1)(x+2)} dx = \int \frac{1}{x+1} dx - \int \frac{1}{x+2} dx = \ln|x+1| - \ln|x+2| + C$

Combine logs:

$\ln\left|\frac{x+1}{x+2}\right| + C$