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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

Take the constant out:

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

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

Apply u-substitution

= - 3 e^{2} \cdot \int e^{u} d u

Use the common integral:

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

= - 3 e^{2} e^{u}

Substitute back

u = e^{x}

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add a constant to the solution

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

\frac{d}{d x} (x 5^{2 x})

\int_{0}^{3} 1

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

\int_{0}^{1} (x^{2} + 1) e^{2 x} d x

t an g e n t 6 x^{2} - 4 x y + 4 y^{2} - 14 = 0, (1, 2)

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