What is the integral of (2000)/(1+0.3e^{-0.276t)} ?

The integral of (2000)/(1+0.3e^{-0.276t)} is 2000(3.62318…ln|1+0.3e^{-0.276t}|+t)+C

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integral of (2000)/(1+0.3e^{-0.276t)}

\int \frac{2000}{1 + 0.3 e ^{- 0.276 t}} d t

2000 (3.62318 \dots ln 1 + 0.3 e^{- 0.276 t} + t) + C

Solution steps

\int \frac{2000}{1 + 0.3 e ^{- 0.276 t}} d t

Take the constant out:

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

= 2000 \cdot \int \frac{1}{1 + 0.3 e ^{- 0.276 t}} d t

\frac{1}{1 + 0.3 e ^{- 0.276 t}} = - \frac{0.3 e ^{- 0.276 t}}{1 + 0.3 e ^{- 0.276 t}} + 1

= 2000 \cdot \int - \frac{0.3 e ^{- 0.276 t}}{1 + 0.3 e ^{- 0.276 t}} + 1 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

= 2000 (- \int \frac{0.3 e ^{- 0.276 t}}{1 + 0.3 e ^{- 0.276 t}} d t + \int 1 d t)

\int \frac{0.3 e ^{- 0.276 t}}{1 + 0.3 e ^{- 0.276 t}} d t = - 3.62318 \dots ln 1 + 0.3 e^{- 0.276 t}

\int 1 d t = t

= 2000 (- (- 3.62318 \dots ln 1 + 0.3 e^{- 0.276 t}) + t)

Simplify

= 2000 (3.62318 \dots ln 1 + 0.3 e^{- 0.276 t} + t)

Add a constant to the solution

= 2000 (3.62318 \dots ln 1 + 0.3 e^{- 0.276 t} + t) + C

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What is the integral of (2000)/(1+0.3e^{-0.276t)} ?
The integral of (2000)/(1+0.3e^{-0.276t)} is 2000(3.62318…ln|1+0.3e^{-0.276t}|+t)+C