What is the integral of x^2e^{x^3-1} ?

The integral of x^2e^{x^3-1} is 1/3 e^{x^3-1}+C

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integral of x^2e^{x^3-1}

\int x^{2} e^{x^{3} - 1} d x

\frac{1}{3} e^{x^{3} - 1} + C

Solution steps

\int x^{2} e^{x^{3} - 1} d x

Apply u-substitution

: \int \frac{e ^{u}}{3} d u

= \int \frac{e ^{u}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{3} e^{u}

Substitute back

u = x^{3} - 1

= \frac{1}{3} e^{x^{3} - 1}

Add a constant to the solution

= \frac{1}{3} e^{x^{3} - 1} + C

\frac{\partial}{\partial x} (x ln (x y - 9))

\frac{4 x ^{2} + 2 x + 8}{x}

x \to \frac{1}{2} lim (10 x^{2} - 9 x + 2)

\int_{1}^{\infty} \frac{1}{x} d x

(0, 2), (- 5, - 1)

What is the integral of x^2e^{x^3-1} ?
The integral of x^2e^{x^3-1} is 1/3 e^{x^3-1}+C