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

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

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integral of sqrt(2+e^{2t)+e^{-2t}}

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

- e^{- t} (- e^{2 t} + 1) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

= \frac{1}{2} \cdot \int - \frac{v + 1}{v v} d v

Take the constant out:

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

= \frac{1}{2} (- \int \frac{v + 1}{v v} d v)

Apply Integration By Parts

= \frac{1}{2} (- (- \frac{2 ( v + 1 )}{v} - \int - \frac{2}{v} d v))

Take the constant out:

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

= \frac{1}{2} (- (- \frac{2 ( v + 1 )}{v} - (- 2 \cdot \int \frac{1}{v} d v)))

Apply the Power Rule

= \frac{1}{2} (- (- \frac{2 ( v + 1 )}{v} - (- 2 \cdot 2 v^{\frac{1}{2}})))

Substitute back

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

Simplify

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

= - e^{- t} (- e^{2 t} + 1)

Add a constant to the solution

= - e^{- t} (- e^{2 t} + 1) + C

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