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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{u^{\frac{3}{2}}} u}{2 x} d u

= 4 \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

= 4 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{3} e^{u}

Substitute back

u = x^{3}

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

Simplify

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

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

Add a constant to the solution

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

y = - 2 x^{3} + \frac{6}{x}, at x = - 1

\int \frac{x ^{3}}{2 x ^{4} + x ^{2} - 1} d x

\frac{d}{d x} (\frac{x ^{4} + 3}{3 x})

\frac{d}{d x} (\frac{x ^{2}}{8 - 8 x})

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

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