What is the integral of e^{x^3}6x^2 ?

The integral of e^{x^3}6x^2 is 2e^{x^3}+C

Popular >

integral of e^{x^3}6x^2

\int e^{x^{3}} 6 x^{2} d x

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

Solution steps

\int e^{x^{3}} \cdot 6 x^{2} d x

Take the constant out:

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

= 6 \cdot \int e^{x^{3}} x^{2} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{3} e^{u}

Substitute back

u = x^{3}

= 6 \cdot \frac{1}{3} e^{x^{3}}

Simplify

6 \cdot \frac{1}{3} e^{x^{3}} : 2 e^{x^{3}}

= 2 e^{x^{3}}

Add a constant to the solution

= 2 e^{x^{3}} + C

\int \frac{2 x + 3}{9 - 4 x ^{2}} d x

\frac{\partial}{\partial x} (5 x^{4} y^{6} + 3 x^{5} y^{8})

x \to 0 lim (\frac{( 6 ^{x} - 4 ^{x} )}{x})

\frac{9 x}{9 x - 5}

(y + 4) \cdot y^{^{'}} = 2 x y + 4 y