What is the integral of (1-t)(8+t^2) ?

The integral of (1-t)(8+t^2) is 8t+(t^3)/3-4t^2-(t^4)/4+C

integral of (1-t)(8+t^2)

\int (1 - t) (8 + t^{2}) d t

8 t + \frac{t ^{3}}{3} - 4 t^{2} - \frac{t ^{4}}{4} + C

Solution steps

\int (1 - t) (8 + t^{2}) d t

Expand

(1 - t) (8 + t^{2}) : 8 + t^{2} - 8 t - t^{3}

= \int 8 + t^{2} - 8 t - t^{3} 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 8 d t + \int t^{2} d t - \int 8 t d t - \int t^{3} d t

\int 8 d t = 8 t

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

\int 8 t d t = 4 t^{2}

\int t^{3} d t = \frac{t ^{4}}{4}

= 8 t + \frac{t ^{3}}{3} - 4 t^{2} - \frac{t ^{4}}{4}

Add a constant to the solution

= 8 t + \frac{t ^{3}}{3} - 4 t^{2} - \frac{t ^{4}}{4} + C

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What is the integral of (1-t)(8+t^2) ?
The integral of (1-t)(8+t^2) is 8t+(t^3)/3-4t^2-(t^4)/4+C