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

The integral of (t^3-1)/(2t-2) is (t^3)/6+(t^2)/4+1/2 t+C

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

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

\frac{t ^{3}}{6} + \frac{t ^{2}}{4} + \frac{1}{2} t + C

Solution steps

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

Expand

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

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

\int \frac{t ^{2}}{2} d t = \frac{t ^{3}}{6}

\int \frac{t}{2} d t = \frac{t ^{2}}{4}

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

= \frac{t ^{3}}{6} + \frac{t ^{2}}{4} + \frac{1}{2} t

Add a constant to the solution

= \frac{t ^{3}}{6} + \frac{t ^{2}}{4} + \frac{1}{2} t + C

What is the integral of (t^3-1)/(2t-2) ?
The integral of (t^3-1)/(2t-2) is (t^3)/6+(t^2)/4+1/2 t+C