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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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 = s - 1

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

Add a constant to the solution

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

\int_{0}^{7} (e^{t} + e^{- t}) d t

x \to 0 + lim ((1 + x)^{\frac{1}{x}})

y = 3 e^{- x}, (0, 3)

\frac{d y}{d x} + \frac{3 x y}{x ^{2} + 1} = \frac{6 x}{x ^{2} + 1}

\int \frac{1}{4 + 8 x} d x

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