What is the integral of (e^{2x})/(e^{5-x)} ?

The integral of (e^{2x})/(e^{5-x)} is 1/3 e^{3x-5}+C

integral of (e^{2x})/(e^{5-x)}

\int \frac{e ^{2 x}}{e ^{5 - x}} d x

\frac{1}{3} e^{3 x - 5} + C

Solution steps

\int \frac{e ^{2 x}}{e ^{5 - x}} d x

Simplify

\frac{e ^{2 x}}{e ^{5 - x}} : e^{3 x - 5}

= \int e^{3 x - 5} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 3 x - 5

= \frac{1}{3} e^{3 x - 5}

Add a constant to the solution

= \frac{1}{3} e^{3 x - 5} + C

x \to \infty lim (67)

\int_{- 2}^{x} \frac{x + 2}{3} d x

d er i v a t i v e \frac{3}{x ^{9}}

x \to \infty lim ((x + 1)^{x + 2})

\frac{d}{d x} (- \frac{81}{( 1 + 9 x ) ^{2}})

What is the integral of (e^{2x})/(e^{5-x)} ?
The integral of (e^{2x})/(e^{5-x)} is 1/3 e^{3x-5}+C