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

To integrate $\frac{1}{(x + 1)(x + 2)}$, decompose into partial fractions:

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

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

$1 = A(x + 2) + B(x + 1)$

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Let $x = -1$:

$1 = A(-1 + 2) + B(0)$

$1 = A$

$A = 1$

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Let $x = -2$:

$1 = A(0) + B(-2 + 1)$

$1 = B(-1)$

$B = -1$

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The fraction becomes:

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

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$\int \frac{1}{(x + 1)(x + 2)} dx = \int \left(\frac{1}{x + 1} - \frac{1}{x + 2}\right) dx$

$= \int \frac{1}{x + 1} dx - \int \frac{1}{x + 2} dx$

$= \log_e|x + 1| - \log_e|x + 2| + C$

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Using the logarithm property $\log a - \log b = \log\left(\frac{a}{b}\right)$:

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