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

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

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

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

t a y l or (1 + \frac{x}{2})^{- 4}

3 x^{2} y^{^{''}} + 7 x y^{^{'}} + y = 24 x + 2

\frac{d}{d x} (\frac{x - 11}{x - 11})

\int \frac{1}{1 + 2 e ^{x} - e ^{- x}} d x

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