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

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

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

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

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

Solution steps

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

Expand

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

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

\int 2 d t = 2 t

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (arccot (\frac{x}{2}))

f (x) = 7 - x^{2}

\frac{d}{d x} (x^{2} - 2 x)

\frac{d}{d x} (\frac{x - 4}{3 x - x ^{2}})

\int 2 + 2 cos (x) d x

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