What is the integral from 0 to x of t(e^{t^3}-1) ?

The integral from 0 to x of t(e^{t^3}-1) is

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integral from 0 to x of t(e^{t^3}-1)

\int_{0}^{x} t (e^{t^{3}} - 1) d t

Solution steps

\int_{0}^{x} t (e^{t^{3}} - 1) d t

Expand

t (e^{t^{3}} - 1) : e^{t^{3}} t - t

= \int_{0}^{x} e^{t^{3}} t - t 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_{0}^{x} e^{t^{3}} t d t - \int_{0}^{x} t d t

\int_{0}^{x} t d t = \frac{x ^{2}}{2}

\int_{1}^{3} 2 x - 3 d x

x (8 - y) = \frac{d y}{d x}

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

\frac{d}{d x} (6 x^{- 3})

x \to \infty lim (y^{x})

What is the integral from 0 to x of t(e^{t^3}-1) ?
The integral from 0 to x of t(e^{t^3}-1) is