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

Answer 1

Elijah Abram

#int (x-1) / (x+1) dx = x -2lnabs(x+1) +C#

Split the integrand function:

#int (x-1) / (x+1) dx = int (x+1-2) / (x+1) dx = int (1-2/(x+1))dx#

Using the linearity of the integral:

#int (x-1) / (x+1) dx = int dx -2 int dx/(x+1)#

These are standard integrals that we can solve directly:

#int (x-1) / (x+1) dx = x -2lnabs(x+1) +C#

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Answer 2

Emma Aldridge

To evaluate the integral (\int \frac{x - 1}{x + 1} , dx), you can use substitution method.

Let (u = x + 1), then (du = dx).

Substitute: [ \int \frac{x - 1}{x + 1} , dx = \int \frac{u - 2}{u} , du ]

Now, divide: [ \int \frac{u - 2}{u} , du = \int \left(1 - \frac{2}{u}\right) , du ]

Integrate: [ \int \left(1 - \frac{2}{u}\right) , du = \int 1 , du - \int \frac{2}{u} , du ]

[ = u - 2 \ln|u| + C ]

Substitute back: [ = (x + 1) - 2 \ln|x + 1| + 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.