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

The integral of 1/((e^{2x)+4)} is 1/8 (-ln|e^{2x}+4|+2x)+C

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

\int \frac{1}{( e ^{2 x} + 4 )} d x

\frac{1}{8} (- ln e^{2 x} + 4 + 2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ( u - 4 )} d u

= \int \frac{1}{2 u ( u - 4 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ( u - 4 )} d u

Take the partial fraction of

\frac{1}{u ( u - 4 )} : - \frac{1}{4 u} + \frac{1}{4 ( u - 4 )}

= \frac{1}{2} \cdot \int - \frac{1}{4 u} + \frac{1}{4 ( u - 4 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int \frac{1}{4 u} d u + \int \frac{1}{4 ( u - 4 )} d u)

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u - 4 )} d u = \frac{1}{4} ln ∣ u - 4 ∣

= \frac{1}{2} (- \frac{1}{4} ln ∣ u ∣ + \frac{1}{4} ln ∣ u - 4 ∣)

Substitute back

u = e^{2 x} + 4

= \frac{1}{2} (- \frac{1}{4} ln e^{2 x} + 4 + \frac{1}{4} ln e^{2 x} + 4 - 4)

Simplify

\frac{1}{2} (- \frac{1}{4} ln e^{2 x} + 4 + \frac{1}{4} ln e^{2 x} + 4 - 4) : \frac{1}{8} (- ln e^{2 x} + 4 + 2 x)

= \frac{1}{8} (- ln e^{2 x} + 4 + 2 x)

Add a constant to the solution

= \frac{1}{8} (- ln e^{2 x} + 4 + 2 x) + C

\frac{\partial}{\partial y} (2 x y + y^{4})

\int \frac{- x ^{2}}{2} d x

\int_{- π}^{π} x^{2} d x

x \to 4 lim (\frac{x}{4 - x} - x^{3})

y^{^{''}} + 8 y^{^{'}} + 15 y = cos (x)

What is the integral of 1/((e^{2x)+4)} ?
The integral of 1/((e^{2x)+4)} is 1/8 (-ln|e^{2x}+4|+2x)+C