What is the integral of (t^3+2t^2-2t)/(t^3-t^2) ?

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

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integral of (t^3+2t^2-2t)/(t^3-t^2)

\int \frac{t ^{3} + 2 t ^{2} - 2 t}{t ^{3} - t ^{2}} d t

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

Solution steps

\int \frac{t ^{3} + 2 t ^{2} - 2 t}{t ^{3} - t ^{2}} d t

Expand

\frac{t ^{3} + 2 t ^{2} - 2 t}{t ^{3} - t ^{2}} : \frac{t ^{3}}{t ^{3} - t ^{2}} + \frac{2 t ^{2}}{t ^{3} - t ^{2}} - \frac{2 t}{t ^{3} - t ^{2}}

Apply the Sum Rule:

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

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

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

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

\int \frac{2 t}{t ^{3} - t ^{2}} d t = 2 (- ln ∣ t ∣ + ln ∣ t - 1 ∣)

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

Simplify

\frac{t ^{2}}{t - 1} + ln ∣ t - 1 ∣ - \frac{1}{t - 1} + 2 ln ∣ t - 1 ∣ - 2 (- ln ∣ t ∣ + ln ∣ t - 1 ∣) : \frac{t ^{2}}{t - 1} + ln ∣ t - 1 ∣ - \frac{1}{t - 1} + 2 ln ∣ t ∣

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

Add a constant to the solution

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

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