What is the integral from 0 to x of sqrt(e^t-1) ?

The integral from 0 to x of sqrt(e^t-1) is 2(-arctan(sqrt(e^x-1))+sqrt(e^x-1))

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integral from 0 to x of sqrt(e^t-1)

\int_{0}^{x} e^{t} - 1 d t

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

Solution steps

\int_{0}^{x} e^{t} - 1 d t

Apply u-substitution

: \int_{1}^{e^{x}} \frac{u - 1}{u} d u

= \int_{1}^{e^{x}} \frac{u - 1}{u} d u

Apply u-substitution

: \int_{0}^{e^{x} - 1} \frac{2 v ^{2}}{v ^{2} + 1} d v

= \int_{0}^{e^{x} - 1} \frac{2 v ^{2}}{v ^{2} + 1} d v

Take the constant out:

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

= 2 \cdot \int_{0}^{e^{x} - 1} \frac{v ^{2}}{v ^{2} + 1} d v

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

= 2 \cdot \int_{0}^{e^{x} - 1} - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

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

= 2 (- \int_{0}^{e^{x} - 1} \frac{1}{v ^{2} + 1} d v + \int_{0}^{e^{x} - 1} 1 d v)

\int_{0}^{e^{x} - 1} \frac{1}{v ^{2} + 1} d v = arctan (e^{x} - 1)

\int_{0}^{e^{x} - 1} 1 d v = e^{x} - 1

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

y = 18 x, (9, 54)

\frac{\partial}{\partial x} (y ln (x y^{2} z^{3}))

\int (1)^{2} d x

y^{^{'}} = 3 cos (t) - 2 y, y (0) = 0

y^{^{'}} + 9 y = t + e - 8 t

What is the integral from 0 to x of sqrt(e^t-1) ?
The integral from 0 to x of sqrt(e^t-1) is 2(-arctan(sqrt(e^x-1))+sqrt(e^x-1))