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

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

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

\int_{0}^{l n (x)} e^{x} - 1 d x

2 (- arctan (x - 1) + x - 1)

Solution steps

\int_{0}^{l n (x)} e^{x} - 1 d x

Apply u-substitution

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

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

Apply u-substitution

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

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

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

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

= 2 (- arctan (x - 1) + x - 1)

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