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

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

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

integral

(e^{x - m})^{2}

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

Solution steps

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

(e^{x - m})^{2} = (e^{- m})^{2} (e^{x})^{2}

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

Take the constant out:

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

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

Simplify

(e^{- m})^{2} : e^{- 2 m}

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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