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

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

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

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

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

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \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

\int_{l n (4)}^{l n (25)} e^{\frac{x}{2}} d x

x \to 0 lim (- \frac{5}{1 - 5 x})

\frac{d y}{d x} = \frac{1}{1 - x}, y (0) = 1

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

x^{3} - x^{2} + 1

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