What is the integral of (e^{2x})/(-2) ?

The integral of (e^{2x})/(-2) is -1/4 e^{2x}+C

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integral of (e^{2x})/(-2)

\int \frac{e ^{2 x}}{- 2} d x

- \frac{1}{4} e^{2 x} + C

Solution steps

\int \frac{e ^{2 x}}{- 2} d x

Take the constant out:

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

= \frac{1}{- 2} \cdot \int e^{2 x} d x

Simplify

= (- \frac{1}{2}) \cdot \int e^{2 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= (- \frac{1}{2}) \cdot \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= (- \frac{1}{2}) \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= (- \frac{1}{2}) \frac{1}{2} e^{u}

Substitute back

u = 2 x

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

Simplify

(- \frac{1}{2}) \frac{1}{2} e^{2 x} : - \frac{1}{4} e^{2 x}

= - \frac{1}{4} e^{2 x}

Add a constant to the solution

= - \frac{1}{4} e^{2 x} + C

\int_{0}^{\infty} \frac{1}{9 + x ^{2}} d x

\int \frac{1}{( x ^{2} + 2 x )} d x

x^{2} - x^{3} + 4

x \to \infty lim (x^{3} - 1)

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

What is the integral of (e^{2x})/(-2) ?
The integral of (e^{2x})/(-2) is -1/4 e^{2x}+C