What is the integral of x^2e^{x^3+4} ?

The integral of x^2e^{x^3+4} is 1/3 e^{x^3+4}+C

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integral of x^2e^{x^3+4}

\int x^{2} e^{x^{3} + 4} d x

\frac{1}{3} e^{x^{3} + 4} + C

Solution steps

\int x^{2} e^{x^{3} + 4} 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} + 4

= \frac{1}{3} e^{x^{3} + 4}

Add a constant to the solution

= \frac{1}{3} e^{x^{3} + 4} + C

f (x) = \frac{( x ^{2} - 1 )}{( x ^{2} + 1 )}

\frac{x + 9}{x - 9}

\int (\frac{2}{x} - \frac{1}{2}) d x

\int x 9 1 + x^{2} d x

\frac{\partial}{\partial x} (3 x + y + e^{x} + 2 e^{y})

What is the integral of x^2e^{x^3+4} ?
The integral of x^2e^{x^3+4} is 1/3 e^{x^3+4}+C