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

Answer 1

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

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

#A/1 + B/(1 + sqrt(x)) = (1 - sqrt(x))/(1 + sqrt(x))#
#A(1 + sqrt(x)) + B = 1 - sqrt(x)#
#A + Asqrt(x) + B = 1 - sqrt(x)#
This means that #{(A = -1), (A + B = 1):}#
Solving, we get that #A = -1# and #B = 2#. The integral becomes
#int -1 + 2/(1 + sqrt(x))dx#
#int -1dx + int 2/(1 + sqrt(x))dx#
#int -1dx + 2int 1/(1 + sqrt(x))dx#
Let #u = 1 + sqrt(x)#. Then #du = 1/(2sqrt(x))dx# and #dx = 2sqrt(x)du#. We can also conclude that #2sqrt(x) = 2(u - 1)#, because #sqrt(x) = u -1# and we have two of those.
#int -1dx + 2int 1/u * 2(u - 1)du#
#int -1dx + 4int (u - 1)/u du#
#int -1dx + 4int u/udu - 4int 1/udu#
#int -1dx + 4int 1du- 4int 1/u du#
#-x + 4u - 4ln|u| + C#
#-x + 4(1+ sqrt(x)) - 4ln|1 + sqrt(x)| + C#

Hopefully this helps!

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Answer 2

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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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.

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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