What is the integral of sqrt((e^x)^2-1) ?

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

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integral of sqrt((e^x)^2-1)

\int (e^{x})^{2} - 1 d x

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

Solution steps

\int (e^{x})^{2} - 1 d x

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= \int e^{2 x} - 1 d x

Apply u-substitution

: \int \frac{e ^{u} - 1}{2} d u

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{2 w ^{2}}{w ^{2} + 1} d w

= \frac{1}{2} \cdot \int \frac{2 w ^{2}}{w ^{2} + 1} d w

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{w ^{2}}{w ^{2} + 1} d w

\frac{w ^{2}}{w ^{2} + 1} = - \frac{1}{w ^{2} + 1} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

\int 1 d w = w

= \frac{1}{2} \cdot 2 (- arctan (w) + w)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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