What is the integral of (1+3t)t^2 ?

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

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integral of (1+3t)t^2

\int (1 + 3 t) t^{2} d t

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

Solution steps

\int (1 + 3 t) t^{2} d t

Expand

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

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

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

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

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

Add a constant to the solution

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

x \to \infty lim (x e^{- x})

\frac{d}{d x} (tan (0))

\frac{\partial}{\partial x} (y^{3})

in (sin (x))

\frac{d}{d x} (\frac{1}{csc ^{2} ( x )})

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