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

#-x + 4(1+ sqrt(x)) - 4ln|1 + sqrt(x)| + C#

Use partial fractions to get rid of the radical in the numerator.

Hopefully this helps!

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To evaluate the integral (\int \frac{1 - \sqrt{x}}{1 + \sqrt{x}} , dx), we can use a substitution method. Let (u = \sqrt{x}), then (du = \frac{1}{2\sqrt{x}} , dx), which implies (2\sqrt{x} , du = dx).

Substitute (u = \sqrt{x}) and (dx = 2\sqrt{x} , du) into the integral:

(\int \frac{1 - \sqrt{x}}{1 + \sqrt{x}} , dx = \int \frac{1 - u}{1 + u} \cdot 2u , du).

This simplifies to:

(\int \frac{(1 - u)2u}{1 + u} , du = \int \frac{2u - 2u^2}{1 + u} , du).

Now, perform the long division:

(2u - 2u^2 = (2u - 2) + \frac{-2u^2 + 2u}{1 + u} = (2u - 2) + \frac{-2u(u - 1)}{1 + u}).

So, (\int \frac{2u - 2u^2}{1 + u} , du = \int (2 - \frac{2u}{1 + u}) , du).

Integrate term by term:

(\int (2 - \frac{2u}{1 + u}) , du = 2u - 2\ln|1 + u| + C).

Replace (u) with (\sqrt{x}):

(= 2\sqrt{x} - 2\ln|1 + \sqrt{x}| + C).

Therefore, the integral (\int \frac{1 - \sqrt{x}}{1 + \sqrt{x}} , dx) evaluates to (2\sqrt{x} - 2\ln|1 + \sqrt{x}| + C), where (C) is the constant of integration.

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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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