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

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Answer 1 · 2017-02-28T00:48:59

It is

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

Answer 2 · 2017-02-28T00:53:53

#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#