# What is the arc length of #f(x)=lnx # in the interval #[1,5]#?

The arc length is

Arc length is given by:

Rearrange:

Rewrite as:

Hence

Integrate term by term:

Reverse the substitution:

Insert the limits of integration:

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To find the arc length of the function ( f(x) = \ln(x) ) in the interval [1, 5], we use the formula for arc length:

[ L = \int_a^b \sqrt{1 + [f'(x)]^2} , dx ]

where ( a = 1 ), ( b = 5 ), and ( f'(x) ) is the derivative of ( f(x) ).

The derivative of ( f(x) = \ln(x) ) is ( f'(x) = \frac{1}{x} ).

Substitute the values into the formula:

[ L = \int_1^5 \sqrt{1 + \left(\frac{1}{x}\right)^2} , dx ]

[ L = \int_1^5 \sqrt{1 + \frac{1}{x^2}} , dx ]

This integral can be solved using trigonometric substitution. Let ( u = \frac{1}{x} ), then ( du = -\frac{1}{x^2} , dx ).

[ L = \int \sqrt{1 + u^2} , (-du) ]

[ L = -\int \sqrt{1 + u^2} , du ]

Now we integrate with respect to ( u ):

[ L = -\frac{1}{2} \left(u \sqrt{1 + u^2} + \ln\left(u + \sqrt{1 + u^2}\right)\right) + C ]

[ L = -\frac{1}{2} \left(\frac{1}{x} \sqrt{1 + \frac{1}{x^2}} + \ln\left(\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}\right)\right) + C ]

Evaluate this expression from ( x = 1 ) to ( x = 5 ) to find the arc length.

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