What is the integral of (t^3)/(t^2-4t+4) ?

The integral of (t^3)/(t^2-4t+4) is (t^2)/2+4t-(12t-16)/(t-2)+12ln|t-2|+C

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integral of (t^3)/(t^2-4t+4)

\int \frac{t ^{3}}{t ^{2} - 4 t + 4} d t

\frac{t ^{2}}{2} + 4 t - \frac{12 t - 16}{t - 2} + 12 ln ∣ t - 2 ∣ + C

Solution steps

\int \frac{t ^{3}}{t ^{2} - 4 t + 4} d t

Long division

\frac{t ^{3}}{t ^{2} - 4 t + 4} : t + 4 + \frac{12 t - 16}{t ^{2} - 4 t + 4}

= \int t + 4 + \frac{12 t - 16}{t ^{2} - 4 t + 4} 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 d t + \int 4 d t + \int \frac{12 t - 16}{t ^{2} - 4 t + 4} d t

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

\int 4 d t = 4 t

\int \frac{12 t - 16}{t ^{2} - 4 t + 4} d t = - \frac{12 t - 16}{t - 2} + 12 ln ∣ t - 2 ∣

= \frac{t ^{2}}{2} + 4 t - \frac{12 t - 16}{t - 2} + 12 ln ∣ t - 2 ∣

Add a constant to the solution

= \frac{t ^{2}}{2} + 4 t - \frac{12 t - 16}{t - 2} + 12 ln ∣ t - 2 ∣ + C

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What is the integral of (t^3)/(t^2-4t+4) ?
The integral of (t^3)/(t^2-4t+4) is (t^2)/2+4t-(12t-16)/(t-2)+12ln|t-2|+C