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

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

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integral of e^{-2t}+2

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

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

Solution steps

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

Apply the Sum Rule:

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

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

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

\int 2 d t = 2 t

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

Add a constant to the solution

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

f (x) = \frac{1}{2} (7 x^{2} + x)^{- 3}

\int_{0}^{1} \int_{4 y}^{4} e^{x^{2}} d x d y

\int \frac{x ^{2}}{( x + 3 ) ^{3}} d x

\frac{d v}{d t} + \frac{( v ^{^{'}} \cdot v )}{t} = v^{^{'}} \cdot (\frac{t}{2} - \frac{1}{2})

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