What is the integral of (1-t)(7+t^2) ?

The integral of (1-t)(7+t^2) is 7t+(t^3)/3-(7t^2)/2-(t^4)/4+C

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integral of (1-t)(7+t^2)

\int (1 - t) (7 + t^{2}) d t

7 t + \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} - \frac{t ^{4}}{4} + C

Solution steps

\int (1 - t) (7 + t^{2}) d t

Expand

(1 - t) (7 + t^{2}) : 7 + t^{2} - 7 t - t^{3}

= \int 7 + t^{2} - 7 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 7 d t + \int t^{2} d t - \int 7 t d t - \int t^{3} d t

\int 7 d t = 7 t

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

\int 7 t d t = \frac{7 t ^{2}}{2}

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

= 7 t + \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} - \frac{t ^{4}}{4}

Add a constant to the solution

= 7 t + \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} - \frac{t ^{4}}{4} + C

What is the integral of (1-t)(7+t^2) ?
The integral of (1-t)(7+t^2) is 7t+(t^3)/3-(7t^2)/2-(t^4)/4+C