int frac{dx}{(x + 1)(x + 2)} = | vedclass

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(C) To evaluate the integral $\int \frac{dx}{(x + 1)(x + 2)}$,we use partial fractions.We can write the integrand as:$\frac{1}{(x + 1)(x + 2)} = \frac

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

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(C) To evaluate the integral $\int \frac{dx}{(x + 1)(x + 2)}$,we use partial fractions.We can write the integrand as:$\frac{1}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$Multiplying both sides by $(x + 1)(x + 2)$,we get:$1 = A(x + 2) + B(x + 1)$Setting $x = -1$,we get $A = 1$.Setting $x = -2$,we get $B = -1$.Thus,$\int \frac{dx}{(x + 1)(x + 2)} = \int \left( \frac{1}{x + 1} - \frac{1}{x + 2} \right) dx$$= \log |x + 1| - \log |x + 2| + C$$= \log \left| \frac{x + 1}{x + 2} \right| + C$.

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

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