How do you evaluate the integral #int lnabs(x-5)dx# from 0 to 2 if it converges?

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Answer 1 · 2017-03-18T14:10:37

#int_0^2 ln abs (x-5) dx =5ln 5 -3 ln3 - 2 #

Evaluate the indefinite integral by parts, considering that:

#d/dx ln abs (x-5) = 1/(x-5) #

so:

#int ln abs (x-5) dx = xlnabs(x-5) - int x/(x-5) dx#

#int ln abs (x-5) dx = xlnabs(x-5) - int (x-5+5)/(x-5) dx#

#int ln abs (x-5) dx = xlnabs(x-5) - int dx -5int dx/(x-5)#

#int ln abs (x-5) dx = xlnabs(x-5) - x -5ln abs(x-5) +C#

#int ln abs (x-5) dx = (x-5)lnabs(x-5) - x + C#

Now we have:

#int_0^2 ln abs (x-5) dx = (2-5)lnabs(2-5) - 2 - (0-5)ln abs (0-5) #

#int_0^2 ln abs (x-5) dx = -3 ln3 - 2 +5ln 5#