What is the integral of (4x^2-8x+4)/(2x-2) ?

The integral of (4x^2-8x+4)/(2x-2) is x^2-2x+C

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integral of (4x^2-8x+4)/(2x-2)

\int \frac{4 x ^{2} - 8 x + 4}{2 x - 2} d x

x^{2} - 2 x + C

Solution steps

\int \frac{4 x ^{2} - 8 x + 4}{2 x - 2} d x

Long division

\frac{4 x ^{2} - 8 x + 4}{2 x - 2} : 2 x - 2

= \int 2 x - 2 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 x d x - \int 2 d x

\int 2 x d x = x^{2}

\int 2 d x = 2 x

= x^{2} - 2 x

Add a constant to the solution

= x^{2} - 2 x + C

What is the integral of (4x^2-8x+4)/(2x-2) ?
The integral of (4x^2-8x+4)/(2x-2) is x^2-2x+C