What is the integral of (1/(e^{t/4)}-e^{t/4})^2 ?

The integral of (1/(e^{t/4)}-e^{t/4})^2 is t/(e^{t/2)}-e^{-t/2}(t+2)-2t+2e^{t/2}+C

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integral of (1/(e^{t/4)}-e^{t/4})^2

\int (\frac{1}{e ^{\frac{t}{4}}} - e^{\frac{t}{4}})^{2} d t

\frac{t}{e ^{\frac{t}{2}}} - e^{- \frac{t}{2}} (t + 2) - 2 t + 2 e^{\frac{t}{2}} + C

Solution steps

\int (\frac{1}{e ^{\frac{t}{4}}} - e^{\frac{t}{4}})^{2} d t

Simplify

(\frac{1}{e ^{\frac{t}{4}}} - e^{\frac{t}{4}})^{2} : \frac{( 1 - e ^{\frac{t}{2}} ) ^{2}}{e ^{\frac{t}{2}}}

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

Expand

\frac{( 1 - e ^{\frac{t}{2}} ) ^{2}}{e ^{\frac{t}{2}}} : \frac{1}{e ^{\frac{t}{2}}} - 2 + e^{\frac{t}{2}}

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

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

\int 2 d t = 2 t

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

= \frac{t}{e ^{\frac{t}{2}}} - e^{- \frac{t}{2}} (t + 2) - 2 t + 2 e^{\frac{t}{2}}

Add a constant to the solution

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

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What is the integral of (1/(e^{t/4)}-e^{t/4})^2 ?
The integral of (1/(e^{t/4)}-e^{t/4})^2 is t/(e^{t/2)}-e^{-t/2}(t+2)-2t+2e^{t/2}+C