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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - \frac{v + 1}{v v} d v

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

= \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

: 2 v^{\frac{1}{2}}

= \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

\int x ln (x^{2} + 1) d x

\frac{1}{s ^{2} ( s + 1.1 )}

\int \frac{1}{2 + x - x ^{2}} d x

\int x x^{2} - 5 d x

x \to - 1 - lim (arccos (x))

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