What is the integral from 0 to 1 of 1/(e^{2x+1)-1} ?

The integral from 0 to 1 of 1/(e^{2x+1)-1} is (ln(e^3-1)-ln(e-1))/2-1

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integral from 0 to 1 of 1/(e^{2x+1)-1}

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

\frac{ln ( e ^{3} - 1 ) - ln ( e - 1 )}{2} - 1

Decimal Notation

0.20380 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{1}{2 ( e ^{u} - 1 )} d u

= \int_{1}^{3} \frac{1}{2 ( e ^{u} - 1 )} 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_{1}^{3} \frac{1}{e ^{u} - 1} d u

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

= \frac{1}{2} \cdot \int_{1}^{3} \frac{e ^{u}}{e ^{u} - 1} - 1 d u

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} (\int_{1}^{3} \frac{e ^{u}}{e ^{u} - 1} d u - \int_{1}^{3} 1 d u)

\int_{1}^{3} \frac{e ^{u}}{e ^{u} - 1} d u = ln (e^{3} - 1) - ln (e - 1)

\int_{1}^{3} 1 d u = 2

= \frac{1}{2} (ln (e^{3} - 1) - ln (e - 1) - 2)

Simplify

\frac{1}{2} (ln (e^{3} - 1) - ln (e - 1) - 2) : \frac{ln ( e ^{3} - 1 ) - ln ( e - 1 )}{2} - 1

= \frac{ln ( e ^{3} - 1 ) - ln ( e - 1 )}{2} - 1

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What is the integral from 0 to 1 of 1/(e^{2x+1)-1} ?
The integral from 0 to 1 of 1/(e^{2x+1)-1} is (ln(e^3-1)-ln(e-1))/2-1