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

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

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integral of 1/(e^{2t)}

\int \frac{1}{e ^{2 t}} d t

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

Solution steps

\int \frac{1}{e ^{2 t}} d t

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int e^{- 2 t} d t

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 t

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

Add a constant to the solution

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

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

f (x) = x^{2} \cdot e^{x^{2} - 1}, at x = - 1

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

\frac{d}{d h} (35 f (h) (3 + 2 h))

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

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