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

The integral of ((e^3)/(e^{x-1)})^2 is -1/2 e^{2(-x+4)}+C

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integral of ((e^3)/(e^{x-1)})^2

\int (\frac{e ^{3}}{e ^{x - 1}})^{2} d x

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

Solution steps

\int (\frac{e ^{3}}{e ^{x - 1}})^{2} d x

Simplify

(\frac{e ^{3}}{e ^{x - 1}})^{2} : e^{2 (- x + 4)}

= \int e^{2 (- x + 4)} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 (- x + 4)

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

Add a constant to the solution

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

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What is the integral of ((e^3)/(e^{x-1)})^2 ?
The integral of ((e^3)/(e^{x-1)})^2 is -1/2 e^{2(-x+4)}+C