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

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(A) The integrand is a proper rational function. We use the method of partial fractions to decompose it.$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{

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

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(A) The integrand is a proper rational function. We use the method of partial fractions to decompose it.$\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$  ..........$(1)$Multiplying both sides by $(x+1)(x+2)$,we get:$1 = A(x+2) + B(x+1)$To find $A$,set $x = -1$:$1 = A(-1+2) + B(0) \implies A = 1$To find $B$,set $x = -2$:$1 = A(0) + B(-2+1) \implies B = -1$Substituting $A$ and $B$ back into equation $(1)$:$\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}$Now,integrating both sides:$\int \frac{dx}{(x+1)(x+2)} = \int \frac{dx}{x+1} - \int \frac{dx}{x+2}$$= \log |x+1| - \log |x+2| + C$$= \log \left|\frac{x+1}{x+2}\right| + C$

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

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

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