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

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

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integral of (6x^2)(e^{x^3+2})

\int (6 x^{2}) (e^{x^{3} + 2}) d x

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

Solution steps

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

Apply Integration By Parts

: \frac{1}{3} e^{x^{3} + 2} x^{3} - \int e^{x^{3} + 2} x^{5} d x

= 6 (\frac{1}{3} e^{x^{3} + 2} x^{3} - \int e^{x^{3} + 2} x^{5} d x)

\int e^{x^{3} + 2} x^{5} d x = \frac{e ^{2}}{3} (e^{x^{3}} x^{3} - e^{x^{3}})

= 6 (\frac{1}{3} e^{x^{3} + 2} x^{3} - \frac{e ^{2}}{3} (e^{x^{3}} x^{3} - e^{x^{3}}))

Simplify

6 (\frac{1}{3} e^{x^{3} + 2} x^{3} - \frac{e ^{2}}{3} (e^{x^{3}} x^{3} - e^{x^{3}})) : 2 e^{2 + x^{3}}

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

Add a constant to the solution

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

9 cos (x)

y = - x^{2} + 16, y = 0, x = - 2, x = 3

\frac{d y}{d x} = - \frac{3 x ^{2} + 2 y ^{2}}{4 x y}

\int (2 x - 5) (3 x + 1) d x

x \to 0 lim (2 x)

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