What is the integral of 2(3-e^{-t}) ?

The integral of 2(3-e^{-t}) is 2(3t+e^{-t})+C

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integral of 2(3-e^{-t})

\int 2 (3 - e^{- t}) d t

2 (3 t + e^{- t}) + C

Solution steps

\int 2 (3 - e^{- t}) d t

Take the constant out:

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

= 2 \cdot \int 3 - e^{- t} d t

Apply u-substitution

: \int - 3 + e^{u} d u

= 2 \cdot \int - 3 + e^{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int 3 d u + \int e^{u} d u)

\int 3 d u = 3 u

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

= 2 (- 3 u + e^{u})

Substitute back

u = - t

= 2 (- 3 (- t) + e^{- t})

Simplify

= 2 (3 t + e^{- t})

Add a constant to the solution

= 2 (3 t + e^{- t}) + C

\frac{2}{s ^{2} + 4 s + 3}

\frac{d}{d x} (- x^{2} sin (x^{2} y))

x \to 3 lim (x^{3} - 3 x^{2} - 9 x)

\frac{x d y}{d x} - y = x^{2} + y^{2}

\int \frac{z ^{- 3}}{3} d z

What is the integral of 2(3-e^{-t}) ?
The integral of 2(3-e^{-t}) is 2(3t+e^{-t})+C