What is the integral of (2e^{3t})/(1+e^{6t)} ?

The integral of (2e^{3t})/(1+e^{6t)} is 2/3 arctan(e^{3t})+C

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integral of (2e^{3t})/(1+e^{6t)}

\int \frac{2 e ^{3 t}}{1 + e ^{6 t}} d t

\frac{2}{3} arctan (e^{3 t}) + C

Solution steps

\int \frac{2 e ^{3 t}}{1 + e ^{6 t}} d t

Take the constant out:

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

= 2 \cdot \int \frac{e ^{3 t}}{1 + e ^{6 t}} d t

Apply u-substitution

= 2 \cdot \int \frac{1}{6 u u - 1} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{u u - 1} d u

Apply u-substitution

= 2 \cdot \frac{1}{6} \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

= 2 \cdot \frac{1}{6} \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)

= 2 \cdot \frac{1}{6} \cdot 2 arctan (v)

Substitute back

= 2 \cdot \frac{1}{6} \cdot 2 arctan (1 + e^{6 t} - 1)

Simplify

2 \cdot \frac{1}{6} \cdot 2 arctan (1 + e^{6 t} - 1) : \frac{2}{3} arctan (e^{3 t})

= \frac{2}{3} arctan (e^{3 t})

Add a constant to the solution

= \frac{2}{3} arctan (e^{3 t}) + C

What is the integral of (2e^{3t})/(1+e^{6t)} ?
The integral of (2e^{3t})/(1+e^{6t)} is 2/3 arctan(e^{3t})+C