The value of int frac{dx}{(x+1)(x+2)} is | vedclass

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(A) 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+

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

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

The value of $\int \frac{dx}{(x+1)(x+2)}$ is

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