What is the integral of (9e^{4t})/(1+e^{8t)} ?

The integral of (9e^{4t})/(1+e^{8t)} is 9/4 arctan(e^{4t})+C

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integral of (9e^{4t})/(1+e^{8t)}

\int \frac{9 e ^{4 t}}{1 + e ^{8 t}} d t

\frac{9}{4} arctan (e^{4 t}) + C

Solution steps

\int \frac{9 e ^{4 t}}{1 + e ^{8 t}} d t

Take the constant out:

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

= 9 \cdot \int \frac{e ^{4 t}}{1 + e ^{8 t}} d t

Apply u-substitution

: \int \frac{1}{8 u u - 1} d u

= 9 \cdot \int \frac{1}{8 u u - 1} d u

Take the constant out:

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

= 9 \cdot \frac{1}{8} \cdot \int \frac{1}{u u - 1} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 1} d v

= 9 \cdot \frac{1}{8} \cdot \int \frac{2}{v ^{2} + 1} d v

Take the constant out:

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

= 9 \cdot \frac{1}{8} \cdot 2 \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 9 \cdot \frac{1}{8} \cdot 2 arctan (v)

Substitute back

= 9 \cdot \frac{1}{8} \cdot 2 arctan (1 + e^{8 t} - 1)

Simplify

9 \cdot \frac{1}{8} \cdot 2 arctan (1 + e^{8 t} - 1) : \frac{9}{4} arctan (e^{4 t})

= \frac{9}{4} arctan (e^{4 t})

Add a constant to the solution

= \frac{9}{4} arctan (e^{4 t}) + C

x = 1 \sum \infty \frac{6}{x ^{2} + x}

3 + 2 f (x)

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What is the integral of (9e^{4t})/(1+e^{8t)} ?
The integral of (9e^{4t})/(1+e^{8t)} is 9/4 arctan(e^{4t})+C