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

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

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

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

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

Solution steps

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

Complete the square

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

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

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 = t - 1

= - \frac{1}{t - 1}

Add a constant to the solution

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

\frac{1}{x ^{2} + y ^{2}} (x d y - y d x) = 0

cos (2 (t - \frac{5 π}{4}))

x \to 2 + lim (∣ x - 2 ∣)

\int 5 e^{2 x} + \frac{1}{x} d x

\frac{d}{d x} (\frac{1 + csc ( x ^{4} )}{1 - cot ( x ^{4} )})

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