Integrate the rational function: frac{x}{(x+1 | vedclass

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(C) Let $\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.Multiplying both sides by $(x+1)(x+2)$,we get $x = A(x+2) + B(x+1)$.To find $A$,put $x

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$\log \frac{(x+2)^2}{|x+1|} + C$

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(C) Let $\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.Multiplying both sides by $(x+1)(x+2)$,we get $x = A(x+2) + B(x+1)$.To find $A$,put $x = -1$: $-1 = A(-1+2) \Rightarrow A = -1$.To find $B$,put $x = -2$: $-2 = B(-2+1) \Rightarrow -2 = -B \Rightarrow B = 2$.Thus,$\frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}$.Integrating both sides with respect to $x$:$\int \frac{x}{(x+1)(x+2)} dx = \int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) dx$.$= -\log |x+1| + 2\log |x+2| + C$.$= \log |x+2|^2 - \log |x+1| + C$.$= \log \left| \frac{(x+2)^2}{x+1} \right| + C$.

Integrate the rational function: $\frac{x}{(x+1)(x+2)}$

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