What is the integral of 1/(x^2-4x+4) ?

The integral of 1/(x^2-4x+4) is -1/(x-2)+C

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integral of 1/(x^2-4x+4)

\int \frac{1}{x ^{2} - 4 x + 4} d x

- \frac{1}{x - 2} + C

Solution steps

\int \frac{1}{x ^{2} - 4 x + 4} d x

Complete the square

x^{2} - 4 x + 4 : (x - 2)^{2}

= \int \frac{1}{( x - 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x - 2

= - \frac{1}{x - 2}

Add a constant to the solution

= - \frac{1}{x - 2} + C

x \to 9 - lim (\frac{6 x}{x - 9})

\frac{d}{d x} (\frac{e ^{4 x}}{x})

\frac{d}{d t} (e^{a t})

\frac{6 x + 4 x ^{2}}{( 3 + 4 x ) ^{2}}

y^{^{''}} - y = 1

What is the integral of 1/(x^2-4x+4) ?
The integral of 1/(x^2-4x+4) is -1/(x-2)+C