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

The integral from 0 to 1 of 1/(3+e^{-2x)} is 1/3-1/3 ln(2)+1/6 ln(1/(e^2)+3)

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

\int_{0}^{1} \frac{1}{3 + e ^{- 2 x}} d x

\frac{1}{3} - \frac{1}{3} ln (2) + \frac{1}{6} ln (\frac{1}{e ^{2}} + 3)

Solution steps

\int_{0}^{1} \frac{1}{3 + e ^{- 2 x}} d x

Apply u-substitution

: \int_{0}^{- 2} - \frac{1}{2 ( 3 + e ^{u} )} d u

= \int_{0}^{- 2} - \frac{1}{2 ( 3 + e ^{u} )} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{0} - \frac{1}{2 ( 3 + e ^{u} )} 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_{- 2}^{0} \frac{1}{3 + e ^{u}} d u)

Apply u-substitution

: \int_{\frac{1}{e ^{2}}}^{1} \frac{1}{v ( 3 + v )} d v

= - (- \frac{1}{2} \cdot \int_{\frac{1}{e ^{2}}}^{1} \frac{1}{v ( 3 + v )} d v)

Take the partial fraction of

\frac{1}{v ( 3 + v )} : \frac{1}{3 v} - \frac{1}{3 ( v + 3 )}

= - (- \frac{1}{2} \cdot \int_{\frac{1}{e ^{2}}}^{1} \frac{1}{3 v} - \frac{1}{3 ( v + 3 )} 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

= - (- \frac{1}{2} (\int_{\frac{1}{e ^{2}}}^{1} \frac{1}{3 v} d v - \int_{\frac{1}{e ^{2}}}^{1} \frac{1}{3 ( v + 3 )} d v))

\int_{\frac{1}{e ^{2}}}^{1} \frac{1}{3 v} d v = \frac{2}{3}

\int_{\frac{1}{e ^{2}}}^{1} \frac{1}{3 ( v + 3 )} d v = \frac{1}{3} (2 ln (2) - ln (\frac{1}{e ^{2}} + 3))

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

Simplify

- (- \frac{1}{2} (\frac{2}{3} - \frac{1}{3} (2 ln (2) - ln (\frac{1}{e ^{2}} + 3)))) : \frac{1}{3} - \frac{1}{3} ln (2) + \frac{1}{6} ln (\frac{1}{e ^{2}} + 3)

= \frac{1}{3} - \frac{1}{3} ln (2) + \frac{1}{6} ln (\frac{1}{e ^{2}} + 3)

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What is the integral from 0 to 1 of 1/(3+e^{-2x)} ?
The integral from 0 to 1 of 1/(3+e^{-2x)} is 1/3-1/3 ln(2)+1/6 ln(1/(e^2)+3)