What is the integral of 3e^{2x^3+3x^2-9}(x^2+x) ?

The integral of 3e^{2x^3+3x^2-9}(x^2+x) is 1/2 e^{2x^3+3x^2-9}+C

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integral of 3e^{2x^3+3x^2-9}(x^2+x)

\int 3 e^{2 x^{3} + 3 x^{2} - 9} (x^{2} + x) d x

\frac{1}{2} e^{2 x^{3} + 3 x^{2} - 9} + C

Solution steps

\int 3 e^{2 x^{3} + 3 x^{2} - 9} (x^{2} + x) d x

Take the constant out:

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

= 3 \cdot \int e^{2 x^{3} + 3 x^{2} - 9} (x^{2} + x) d x

Apply u-substitution

: \int \frac{e ^{u}}{6} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 x^{3} + 3 x^{2} - 9

= 3 \cdot \frac{1}{6} e^{2 x^{3} + 3 x^{2} - 9}

Simplify

3 \cdot \frac{1}{6} e^{2 x^{3} + 3 x^{2} - 9} : \frac{1}{2} e^{2 x^{3} + 3 x^{2} - 9}

= \frac{1}{2} e^{2 x^{3} + 3 x^{2} - 9}

Add a constant to the solution

= \frac{1}{2} e^{2 x^{3} + 3 x^{2} - 9} + C

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

\frac{\partial}{\partial y} (e^{x^{2} + y^{2} - 2})

3 x^{2} + 2

\int_{4}^{14} 16 + (4 - x)^{2} d x

y^{^{'}} = y + 2 e^{- e}

What is the integral of 3e^{2x^3+3x^2-9}(x^2+x) ?
The integral of 3e^{2x^3+3x^2-9}(x^2+x) is 1/2 e^{2x^3+3x^2-9}+C