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$∫\frac{1}{(x+1)(x+2)}dx=$ |
$log(x+1)(x+2)+C;$ Where C is arbitrary constant of integration $log\frac{x+1}{x+2}+C;$ Where C is arbitrary constant of integration $log\frac{x+2}{x+1}+C;$Where C is arbitrary constant of integration $-log(x+1)(x+2)+C;$ Where C is arbitrary constant of integration |
$log\frac{x+1}{x+2}+C;$ Where C is arbitrary constant of integration |
The correct answer is Option (2) → $\log\frac{x+1}{x+2}+C;$ Where C is arbitrary constant of integration $∫\frac{1}{(x+1)(x+2)}dx=∫\frac{(x+2)-(x+1)}{(x+1)(x+2)}dx$ $=∫\frac{1}{(x+1)}-\frac{1}{(x+2)}dx$ $=\log(x+1)-\log(x+2)+C$ $=\log\left|\frac{(x+1)}{(x+2)}\right|+C$ |
