# What is the net area between #f(x) = 1/sqrt(x^2+2x+1) # and the x-axis over #x in [2, 4 ]#?

The answer is ln(5/3)

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To find the net area between the curve ( f(x) = \frac{1}{\sqrt{x^2 + 2x + 1}} ) and the x-axis over ( x ) in the interval ([2, 4]), we need to compute the definite integral of ( f(x) ) over the given interval. The net area can be calculated using the formula for the definite integral of a function.

[ A = \int_{a}^{b} |f(x)| , dx ]

where ( a = 2 ) and ( b = 4 ) in this case.

[ A = \int_{2}^{4} \frac{1}{\sqrt{x^2 + 2x + 1}} , dx ]

This integral can be challenging to evaluate directly due to the square root in the denominator. However, we can simplify it by completing the square in the denominator:

[ x^2 + 2x + 1 = (x + 1)^2 ]

Substituting this expression back into the integral:

[ A = \int_{2}^{4} \frac{1}{|x + 1|} , dx ]

Now, we can split the integral into two parts, as the expression inside the absolute value changes sign at ( x = -1 ):

[ A = \int_{2}^{-1} \frac{1}{-(x + 1)} , dx + \int_{-1}^{4} \frac{1}{x + 1} , dx ]

After evaluating these integrals, the net area between the curve and the x-axis over the interval ([2, 4]) can be found.

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