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

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

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

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

\frac{x ^{2}}{2} - 2 x + C

Solution steps

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

Simplify

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

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

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

\int 2 d x = 2 x

= \frac{x ^{2}}{2} - 2 x

Add a constant to the solution

= \frac{x ^{2}}{2} - 2 x + C

y^{^{''}} = 8 (y^{^{'}})^{2}

y = e^{x}, y = e^{- 2 x}, x = ln (3)

\frac{d}{d x} ((cos (x))^{2} - 2 sin (x))

y = \frac{6}{sin ( x ) + cos ( x )}, (0, 6)

x \to 1 + lim (\frac{4}{2 x - 2})

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