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

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

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

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

2 (ln (e + 1) - \frac{1}{e} - ln (2))

Decimal

0.50447 \dots

Solution steps

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

Apply u-substitution

= \int_{0}^{1} \frac{2}{e ^{u} + e ^{2 u}} d u

Take the constant out:

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

= 2 \cdot \int_{0}^{1} \frac{1}{e ^{u} + e ^{2 u}} d u

Apply u-substitution

= 2 \cdot \int_{1}^{e} \frac{1}{v ( v + v ^{2} )} d v

Take the partial fraction of

\frac{1}{v ( v + v ^{2} )} : - \frac{1}{v} + \frac{1}{v ^{2}} + \frac{1}{v + 1}

= 2 \cdot \int_{1}^{e} - \frac{1}{v} + \frac{1}{v ^{2}} + \frac{1}{v + 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_{1}^{e} \frac{1}{v} d v + \int_{1}^{e} \frac{1}{v ^{2}} d v + \int_{1}^{e} \frac{1}{v + 1} d v)

\int_{1}^{e} \frac{1}{v} d v = 1

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

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

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

Simplify

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

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