What is the integral from 0 to 1 of (e^x)/(16-e^{2x)} ?

The integral from 0 to 1 of (e^x)/(16-e^{2x)} is (ln((3(e+4))/(5(-e+4))))/8

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integral from 0 to 1 of (e^x)/(16-e^{2x)}

\int_{0}^{1} \frac{e ^{x}}{16 - e ^{2 x}} d x

\frac{ln ( \frac{3 ( e + 4 )}{5 ( - e + 4 )} )}{8}

Decimal Notation

0.14322 \dots

Solution steps

\int_{0}^{1} \frac{e ^{x}}{16 - e ^{2 x}} d x

Apply u-substitution

: \int_{1}^{e} \frac{1}{16 - u ^{2}} d u

= \int_{1}^{e} \frac{1}{16 - u ^{2}} d u

Apply Integral Substitution

= \int_{\frac{1}{4}}^{\frac{e}{4}} \frac{1}{4 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int_{\frac{1}{4}}^{\frac{e}{4}} \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{4} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{e}{4}}

Simplify

\frac{1}{4} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{e}{4}} : \frac{1}{4} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{e}{4}}

= \frac{1}{4} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{e}{4}}

Compute the boundaries:

\frac{ln ( \frac{e}{4} + 1 ) - ln ( - \frac{e}{4} + 1 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{2}

= \frac{1}{4} \cdot \frac{ln ( \frac{e}{4} + 1 ) - ln ( - \frac{e}{4} + 1 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{2}

Simplify

= \frac{ln ( \frac{3 ( e + 4 )}{5 ( - e + 4 )} )}{8}

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What is the integral from 0 to 1 of (e^x)/(16-e^{2x)} ?
The integral from 0 to 1 of (e^x)/(16-e^{2x)} is (ln((3(e+4))/(5(-e+4))))/8