What is the integral of (7e^{2t})/(1+e^{4t)} ?

The integral of (7e^{2t})/(1+e^{4t)} is 7/2 arctan(e^{2t})+C

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integral of (7e^{2t})/(1+e^{4t)}

\int \frac{7 e ^{2 t}}{1 + e ^{4 t}} d t

\frac{7}{2} arctan (e^{2 t}) + C

Solution steps

\int \frac{7 e ^{2 t}}{1 + e ^{4 t}} d t

Take the constant out:

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

= 7 \cdot \int \frac{e ^{2 t}}{1 + e ^{4 t}} d t

Apply u-substitution

: \int \frac{1}{2 ( 1 + u ^{2} )} d u

= 7 \cdot \int \frac{1}{2 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= 7 \cdot \frac{1}{2} arctan (u)

Substitute back

u = e^{2 t}

= 7 \cdot \frac{1}{2} arctan (e^{2 t})

Simplify

7 \cdot \frac{1}{2} arctan (e^{2 t}) : \frac{7}{2} arctan (e^{2 t})

= \frac{7}{2} arctan (e^{2 t})

Add a constant to the solution

= \frac{7}{2} arctan (e^{2 t}) + C

\frac{d}{d x} (x^{3} - 2 x + 4)

3 x^{\frac{4}{3}}

x \to 1 lim (\frac{x ^{2} - 1}{x - 3})

cos (a x + b)

\frac{d}{d x} (\frac{( x ^{2} )}{7 + 6 x})

What is the integral of (7e^{2t})/(1+e^{4t)} ?
The integral of (7e^{2t})/(1+e^{4t)} is 7/2 arctan(e^{2t})+C