Using partial fractions, find f(x) if f'(x)=5/(2x-1)(x-3)
First step: partial fractions 5/(2x-1)(x-3) 5=A(x-3)+B(2x-1) A=0 when x=3, so B=5/(2x3-1)=1 B=0 when x=1/2, so A=5/(0.5-3)=-2 So f'(x)=1/(x-3)-2/(2x-1) Second step: Integration f(x)= (integral)(1/(x-3))dx - 2(integral)(1/(2x-1))dx = ln|x-3| - 2/2ln|2x-1| + C = ln|(x-3)/(2x-1)| + C
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