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Question #36feb

1 Answer

Apr 3, 2017

#int (2x-3)/(x-1) = 2x-ln abs(x-1)+C#

Explanation:

Write the integrand function as:

#(2x-3)/(x-1) = 2( (x-3/2)/(x-1)) = 2( (x-1-1/2)/(x-1)) = 2 - 1/(x-1)#

So:

#int (2x-3)/(x-1) = int (2 - 1/(x-1))dx#

and using the linearity of integrals:

#int (2x-3)/(x-1) = 2 int dx - int dx/(x-1) = 2x-ln abs(x-1)+C#

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