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

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

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

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

\frac{1}{2} e^{2 x^{4}} + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int x^{3} e^{2 x^{4}} d x

Apply u-substitution

: \int \frac{e ^{u}}{8} d u

= 4 \cdot \int \frac{e ^{u}}{8} d u

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{8} e^{u}

Substitute back

u = 2 x^{4}

= 4 \cdot \frac{1}{8} e^{2 x^{4}}

Simplify

4 \cdot \frac{1}{8} e^{2 x^{4}} : \frac{1}{2} e^{2 x^{4}}

= \frac{1}{2} e^{2 x^{4}}

Add a constant to the solution

= \frac{1}{2} e^{2 x^{4}} + C

\frac{d}{d x} ((\frac{x ^{2}}{6 - x})^{5})

\int_{0}^{1} x^{4} - 2 x^{3} + x^{2} d x

\frac{\partial}{\partial x} (9 - 2 x^{2} + 3 y^{2})

\frac{\partial}{\partial x} (cos (2 x + 5 y) \cdot 5)

- 15 ln (x)

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