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

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

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

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

\frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C

Solution steps

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

Simplify

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

= \int t^{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 t d t + \int 1 d t

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

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

\int 1 d t = t

= \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t

Add a constant to the solution

= \frac{t ^{3}}{3} + \frac{t ^{2}}{2} + t + C

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