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

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

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integral of (e^{-t})^2

integral

(e^{- t})^{2}

- \frac{1}{2} e^{- 2 t} + C

Solution steps

\int (e^{- t})^{2} d t

Apply u-substitution

: \int - u d u

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

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

Substitute back

u = e^{- t}

= - \frac{( e ^{- t} ) ^{2}}{2}

Simplify

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

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

Add a constant to the solution

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

2 t + 5

5 x^{2}

sec^{2} (x) tan^{2} (x)

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

\frac{1}{- x}

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