What is the integral of \sqrt[3]{x}e^{\sqrt[3]{8x^4}} ?

The integral of \sqrt[3]{x}e^{\sqrt[3]{8x^4}} is 3/8 e^{2x^{4/3}}+C

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integral of \sqrt[3]{x}e^{\sqrt[3]{8x^4}}

\int 3 x e^{3 8 x^{4}} d x

\frac{3}{8} e^{2 x^{\frac{4}{3}}} + C

Solution steps

\int 3 x e^{3 8 x^{4}} d x

Apply u-substitution

: \int 3 e^{2 u^{4}} u^{3} d u

= \int 3 e^{2 u^{4}} u^{3} d u

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \int \frac{e ^{v}}{8} d v

Take the constant out:

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

= 3 \cdot \frac{1}{8} \cdot \int e^{v} d v

Use the common integral:

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

= 3 \cdot \frac{1}{8} e^{v}

Substitute back

= 3 \cdot \frac{1}{8} e^{2 (3 x)^{4}}

Simplify

3 \cdot \frac{1}{8} e^{2 (3 x)^{4}} : \frac{3}{8} e^{2 x^{\frac{4}{3}}}

= \frac{3}{8} e^{2 x^{\frac{4}{3}}}

Add a constant to the solution

= \frac{3}{8} e^{2 x^{\frac{4}{3}}} + C

n = 1 \sum \infty 3

\int (x - sin (x)) d x

\int (x^{2} + 9 x) cos (x) d x

f (x) = (3 x - 2)^{2} (5 x + 4)

x \to \infty + lim (\frac{x ^{3}}{1 e ^{\frac{x}{4}}})

What is the integral of \sqrt[3]{x}e^{\sqrt[3]{8x^4}} ?
The integral of \sqrt[3]{x}e^{\sqrt[3]{8x^4}} is 3/8 e^{2x^{4/3}}+C