What is the integral of sqrt(4+(e^x-e^{-x))^2} ?

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

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integral of sqrt(4+(e^x-e^{-x))^2}

\int 4 + (e^{x} - e^{- x})^{2} d x

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

Solution steps

\int 4 + (e^{x} - e^{- x})^{2} d x

4 + (e^{x} - e^{- x})^{2} = 4 (1 + \frac{( e ^{x} - e ^{- x} ) ^{2}}{4})

= \int 4 (1 + \frac{( e ^{x} - e ^{- x} ) ^{2}}{4}) d x

Simplify

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

= \int e^{- 2 x} + e^{2 x} + 2 d x

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

Simplify

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

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

Add a constant to the solution

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

f (x) = - \frac{8}{( 1 + 4 x ) ^{2}}

\int (t^{2} + 1) e^{t^{3} + 3 t} d t

\int cos (\frac{π x}{4}) d x

f (x) = x^{4} - 4 x^{2} + 4

\frac{d}{d x} (sin (ln (tan (x))))

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