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

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

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

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

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

Solution steps

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

Complete the square

y^{2} - 2 y + 1 : (y - 1)^{2}

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

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

Add a constant to the solution

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

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