# How do you evaluate the integral #int 1/(1+sqrtx)#?

The answer is

We can perform this integral by substitution

So,

Therefore,

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To evaluate the integral ( \int \frac{1}{1 + \sqrt{x}} ), use the substitution method.

Let's set: [ u = \sqrt{x} ] [ \Rightarrow du = \frac{1}{2\sqrt{x}} dx ]

Now, solve for dx: [ dx = 2u du ]

Substitute these values into the integral:

[ \int \frac{1}{1 + u} \cdot 2u du ]

This simplifies to: [ 2 \int \frac{u}{1 + u} du ]

Split the fraction: [ 2 \int \left( 1 - \frac{1}{1 + u} \right) du ]

Integrate term by term:

[ 2 \left( u - \ln|1 + u| \right) + C ]

Replace u with ( \sqrt{x} ):

[ 2 \left( \sqrt{x} - \ln|1 + \sqrt{x}| \right) + C ]

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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.

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.

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.

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