How do you integrate #int (2x-3)/(x^2-3x+5)# using substitution?

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Answer 1 · 2017-02-17T14:14:02

#int (2x-3)/(x^2-3x+5)dx = ln (x^2-3x+5) + C#

Note that:

#d/dx (x^2-3x+5) = 2x-3#

So if we substitute:

#t = x^2-3x+5#

#dt = (2x-3)dx#

we have:

#int (2x-3)/(x^2-3x+5)dx = int (dt)/t = ln abs(t) +C = ln (x^2-3x+5) + C#