Find $\int \frac{dx}{(x+1)(x+2)}$

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

The correct answer is Option (3) → $\ln \left| \frac{x+1}{x+2} \right| + C$

The integrand is a proper rational function. Therefore, by using the form of partial fraction, we write

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

where, real numbers $A$ and $B$ are to be determined suitably. This gives

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

Equating the coefficients of $x$ and the constant term, we get

$A + B = 0$

and

$2A + B = 1$

Solving these equations, we get $A = 1$ and $B = -1$.

Thus, the integrand is given by

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

Therefore,

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

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

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