What is the integral of e^{x^3}(12x^2) ?

The integral of e^{x^3}(12x^2) is 4e^{x^3}+C

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integral of e^{x^3}(12x^2)

\int e^{x^{3}} (12 x^{2}) d x

4 e^{x^{3}} + C

Solution steps

\int e^{x^{3}} \cdot 12 x^{2} d x

Take the constant out:

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

= 12 \cdot \int e^{x^{3}} x^{2} d x

Apply u-substitution

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

= 12 \cdot \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

= 12 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 12 \cdot \frac{1}{3} e^{u}

Substitute back

u = x^{3}

= 12 \cdot \frac{1}{3} e^{x^{3}}

Simplify

12 \cdot \frac{1}{3} e^{x^{3}} : 4 e^{x^{3}}

= 4 e^{x^{3}}

Add a constant to the solution

= 4 e^{x^{3}} + C

\frac{\partial}{\partial x} (x e^{- \frac{y}{x}})

f (t) = \frac{3}{5} e^{- t} - \frac{22}{5} e^{4 t}

\int (\frac{3}{x ^{2}} - 1) d x

\frac{d y}{d x} = 2 - y

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 2

What is the integral of e^{x^3}(12x^2) ?
The integral of e^{x^3}(12x^2) is 4e^{x^3}+C