What is the integral of (e^{(-t)/6}-e^{t/6})^2 ?

The integral of (e^{(-t)/6}-e^{t/6})^2 is -3e^{-t/3}-2t+3e^{t/3}+C

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integral of (e^{(-t)/6}-e^{t/6})^2

\int (e^{\frac{- t}{6}} - e^{\frac{t}{6}})^{2} d t

- 3 e^{- \frac{t}{3}} - 2 t + 3 e^{\frac{t}{3}} + C

Solution steps

\int (e^{\frac{- t}{6}} - e^{\frac{t}{6}})^{2} d t

Expand

(e^{\frac{- t}{6}} - e^{\frac{t}{6}})^{2} : e^{- \frac{t}{3}} - 2 + e^{\frac{t}{3}}

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

\int e^{- \frac{t}{3}} d t = - 3 e^{- \frac{t}{3}}

\int 2 d t = 2 t

\int e^{\frac{t}{3}} d t = 3 e^{\frac{t}{3}}

= - 3 e^{- \frac{t}{3}} - 2 t + 3 e^{\frac{t}{3}}

Add a constant to the solution

= - 3 e^{- \frac{t}{3}} - 2 t + 3 e^{\frac{t}{3}} + C

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What is the integral of (e^{(-t)/6}-e^{t/6})^2 ?
The integral of (e^{(-t)/6}-e^{t/6})^2 is -3e^{-t/3}-2t+3e^{t/3}+C