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

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

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

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

- \frac{2 t - 1}{t - 1} + 2 ln ∣ t - 1 ∣ + C

Solution steps

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

Apply Integration By Parts

= - \frac{2 t - 1}{t - 1} - \int - \frac{2}{t - 1} d t

\int - \frac{2}{t - 1} d t = - 2 ln ∣ t - 1 ∣

= - \frac{2 t - 1}{t - 1} - (- 2 ln ∣ t - 1 ∣)

Simplify

= - \frac{2 t - 1}{t - 1} + 2 ln ∣ t - 1 ∣

Add a constant to the solution

= - \frac{2 t - 1}{t - 1} + 2 ln ∣ t - 1 ∣ + C

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