What is the integral of (1-2x)^2 dx/x

Question

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Answer 1

Note that (1-2x)^2 = (1-2x)(1-2x) = 1-4x+4x^2. Therefore,

(1-2x)^2/x =(1-4x+x^2)/x =(1/x) - (4x/x)+(4x^2/x)= 1/x - 4 + 4x. As a result,

int((1-2x)^2/x)=int(1/x - 4 + 4x) = int(1/x) - int(4) + int(4x) = ln|x| - 4x + 4(1/2 x^2) +C = ln|x|-4x+2x^2 +C.