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

The integral of t^2+(2t-1)/(t^2-2t+1) is (t^3)/3-(2t-1)/(t-1)+2ln|t-1|+C

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

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

\frac{t ^{3}}{3} - \frac{2 t - 1}{t - 1} + 2 ln ∣ t - 1 ∣ + C

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int t^{2} d t = \frac{t ^{3}}{3}

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

= \frac{t ^{3}}{3} - \frac{2 t - 1}{t - 1} + 2 ln ∣ t - 1 ∣

Add a constant to the solution

= \frac{t ^{3}}{3} - \frac{2 t - 1}{t - 1} + 2 ln ∣ t - 1 ∣ + C

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