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How do you find #int (x+2)/sqrt(-x^2-4x)#?

Answer 1

Christopher Agresta

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Answer 2

Aurora Alvarado

To find the integral of (x + 2)/sqrt(-x^2 - 4x), you can use trigonometric substitution. Let ( t = -x - 2 ). Then ( dt = -dx ). Substituting these into the integral, you get [ \int{\frac{-(t - 2)}{\sqrt{t^2}}} , dt ] Simplify the expression to [ -\int{\frac{t - 2}{\sqrt{t^2}}} , dt ] Now, split the fraction and integrate each part separately. [ -\int{\frac{t}{\sqrt{t^2}}} , dt + 2\int{\frac{1}{\sqrt{t^2}}} , dt ] [ = -\int{\frac{t}{|t|}} , dt + 2\int{\frac{1}{|t|}} , dt ] [ = -\int{sgn(t)} , dt + 2\int{\frac{1}{|t|}} , dt ] The integral of ( sgn(t) ) is ( |t| ) and the integral of ( \frac{1}{|t|} ) is ( ln|t| ). So, the integral becomes [ = -|t| + 2ln|t| + C ] Substituting back ( t = -x - 2 ) gives [ = -|-x - 2| + 2ln|-x - 2| + C ] [ = x + 2 + 2ln|-x - 2| + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.