What is the integral of (9e^{2t})/(1+e^{4t)} ?

The integral of (9e^{2t})/(1+e^{4t)} is 9/2 arctan(e^{2t})+C

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integral of (9e^{2t})/(1+e^{4t)}

\int \frac{9 e ^{2 t}}{1 + e ^{4 t}} d t

\frac{9}{2} arctan (e^{2 t}) + C

Solution steps

\int \frac{9 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

= 9 \cdot \int \frac{e ^{2 t}}{1 + e ^{4 t}} d t

Apply u-substitution

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

= 9 \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

= 9 \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)

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

Substitute back

u = e^{2 t}

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

Simplify

9 \cdot \frac{1}{2} arctan (e^{2 t}) : \frac{9}{2} arctan (e^{2 t})

= \frac{9}{2} arctan (e^{2 t})

Add a constant to the solution

= \frac{9}{2} arctan (e^{2 t}) + C

f (x) = \frac{1}{2} x^{2} + 7 x

\int co tan (x) d x

x \to \frac{π}{3} lim (x cos (x))

y^{^{'}} + 2 y = e^{2 x}

x \to 0 + lim (lo g_{3} (x))

What is the integral of (9e^{2t})/(1+e^{4t)} ?
The integral of (9e^{2t})/(1+e^{4t)} is 9/2 arctan(e^{2t})+C