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

The integral of (t^3-3t)(2t) is 2((t^5)/5-t^3)+C

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

\int (t^{3} - 3 t) (2 t) d t

2 (\frac{t ^{5}}{5} - t^{3}) + C

Solution steps

\int (t^{3} - 3 t) \cdot 2 t d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int (t^{3} - 3 t) t d t

Expand

(t^{3} - 3 t) t : t^{4} - 3 t^{2}

= 2 \cdot \int t^{4} - 3 t^{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

= 2 (\int t^{4} d t - \int 3 t^{2} d t)

\int t^{4} d t = \frac{t ^{5}}{5}

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

= 2 (\frac{t ^{5}}{5} - t^{3})

Add a constant to the solution

= 2 (\frac{t ^{5}}{5} - t^{3}) + C

\frac{d x}{d t} = 8 x t^{5}

\int sin^{2} (x) cos^{3} (x) d x

x y + y^{2} - x^{2} y^{^{'}} = 0

\frac{d}{d x} (5^{x + 2})

\frac{d}{d x} (\frac{2}{7 x ^{4}})

What is the integral of (t^3-3t)(2t) ?
The integral of (t^3-3t)(2t) is 2((t^5)/5-t^3)+C