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

The integral of (1-t+2t^2-t^3)/(t^3+2t^2+t) is -2t+ln|t|+3ln|t+1|+4/(t+1)+(t^2)/(t+1)-2+C

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integral of (1-t+2t^2-t^3)/(t^3+2t^2+t)

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

- 2 t + ln ∣ t ∣ + 3 ln ∣ t + 1 ∣ + \frac{4}{t + 1} + \frac{t ^{2}}{t + 1} - 2 + C

Solution steps

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

Expand

\frac{1 - t + 2 t ^{2} - t ^{3}}{t ^{3} + 2 t ^{2} + t} : \frac{1}{t ^{3} + 2 t ^{2} + t} - \frac{t}{t ^{3} + 2 t ^{2} + t} + \frac{2 t ^{2}}{t ^{3} + 2 t ^{2} + t} - \frac{t ^{3}}{t ^{3} + 2 t ^{2} + 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 \frac{1}{t ^{3} + 2 t ^{2} + t} d t - \int \frac{t}{t ^{3} + 2 t ^{2} + t} d t + \int \frac{2 t ^{2}}{t ^{3} + 2 t ^{2} + t} d t - \int \frac{t ^{3}}{t ^{3} + 2 t ^{2} + t} d t

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

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

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

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

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

Simplify

ln ∣ t ∣ - ln ∣ t + 1 ∣ + \frac{1}{t + 1} - (- \frac{1}{t + 1}) + 2 (ln ∣ t + 1 ∣ + \frac{1}{t + 1}) - (- \frac{t ^{2}}{t + 1} + 2 (t + 1 - ln ∣ t + 1 ∣)) : - 2 t + ln ∣ t ∣ + 3 ln ∣ t + 1 ∣ + \frac{4}{t + 1} + \frac{t ^{2}}{t + 1} - 2

= - 2 t + ln ∣ t ∣ + 3 ln ∣ t + 1 ∣ + \frac{4}{t + 1} + \frac{t ^{2}}{t + 1} - 2

Add a constant to the solution

= - 2 t + ln ∣ t ∣ + 3 ln ∣ t + 1 ∣ + \frac{4}{t + 1} + \frac{t ^{2}}{t + 1} - 2 + C

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What is the integral of (1-t+2t^2-t^3)/(t^3+2t^2+t) ?
The integral of (1-t+2t^2-t^3)/(t^3+2t^2+t) is -2t+ln|t|+3ln|t+1|+4/(t+1)+(t^2)/(t+1)-2+C