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

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

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

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

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

Solution steps

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

Complete the square

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

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

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 = y - 2

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

Add a constant to the solution

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

\int (e^{- x} - e^{5}) \cdot (e^{x} + e^{- 3}) d x

\frac{\partial}{\partial x} (y e^{x^{2}})

\frac{\partial}{\partial x} (x^{2} + 3 x y^{2})

\frac{2 s - 7}{( s - 4 ) ( s + 6 )}

\int \frac{x + 4}{x ^{2} + 5 x - 6} d x

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