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How do you integrate #x/(x-6)#?

Answer 1

Scarlett Allison

It is

#int (x/(x-6))dx=int (1+6/(x-6))dx=x+6*lnabs(x-6)+c#

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Answer 2

Charles Arocha

#int x/(x-6)dx =x+6lnabs(x-6)+C#

Complete the numerator to separate the function in a polynomial plus a proper rational function:

#x/(x-6) = (x-6+6)/(x-6) = 1+6/(x-6)#

We can now integrate the two terms separately, using linearity:

#int x/(x-6)dx = int dx+6int dx/(x-6)#

#int x/(x-6)dx =x+6lnabs(x-6)+C#

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Answer 3

Scarlett Allison

To integrate ( \frac{x}{x-6} ), use substitution. Let ( u = x - 6 ), then ( du = dx ). After substitution, integrate ( \frac{u + 6}{u} ) with respect to ( u ). This gives ( \int \left(1 + \frac{6}{u}\right) du ). The integral becomes ( u + 6 \ln|u| + C ), where ( C ) is the constant of integration. Substitute back ( u = x - 6 ) to get the final answer: ( x - 6 + 6 \ln|x - 6| + C ).

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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.