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

The integral of 1/((x^2-2x+1)^2) is -1/(3(x-1)^3)+C

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

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

- \frac{1}{3 ( x - 1 ) ^{3}} + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x ^{2} - 2 x + 1 ) ^{2}} : \frac{1}{( x - 1 ) ^{4}}

= \int \frac{1}{( x - 1 ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{u ^{4}} d u

= \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= - \frac{1}{3 u ^{3}}

Substitute back

u = x - 1

= - \frac{1}{3 ( x - 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{3 ( x - 1 ) ^{3}} + C

y = - x^{3}, (2, - 8)

\frac{d y}{d x} = - (1 + y)^{4}

y^{^{''}} = sin (3 x - \frac{π}{3})

7 e^{- 4 x}

\int \frac{( e ^{x} + 9 )}{e ^{2 x} - 6 e ^{x} + 9} d x

What is the integral of 1/((x^2-2x+1)^2) ?
The integral of 1/((x^2-2x+1)^2) is -1/(3(x-1)^3)+C