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

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

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

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

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

Solution steps

\int 6 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

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

Apply u-substitution

: \int \frac{e ^{u^{\frac{3}{2}}} u}{2 x} 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{x ^{2} + x + 1}{( x + 1 ) ^{2} ( x + 2 )} d x

\frac{d}{d x} ((x cos (x)) (cos (x) + tan (x)))

\int_{0}^{1} 1 - x^{\frac{2}{7}} d x

x^{2} - 3 x y + x y^{^{'}} = 0

\frac{d}{d t} (cos (t))

What is the integral of 6x^2*e^{x^3} ?
The integral of 6x^2*e^{x^3} is 2e^{x^3}+C