How do you evaluate the definite integral #int dx/(1-x)# from #[-2,0]#?

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Answer 1 · 2018-05-01T02:11:12

# int_(-2)^0 \ 1/(1-x) \ dx ln 3 #

We seek:

# I = int_(-2)^0 \ 1/(1-x) \ dx #

Noting that the integrand is continuous or over the range of integration, we can apply a simply substituting, Let

# u-1-x => (du)/dx = -1 #

And we have a transformation of the integration limits:

# x = { (-2),(0) :} => u = { (3),(1) :} #

Thus we have:

# I = int_(3)^1 \ 1/u \ (-1) \ du #

# \ \ = int_(1)^3 \ 1/u \ du #

# \ \ = [lnu]_(1)^3 #

# \ \ = ln 3 - ln 1 #

# \ \ = ln 3 #