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

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

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integral of 1/(e^x+4)

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

\frac{1}{4} x - \frac{1}{4} ln ∣ e^{x} + 4 ∣ + C

Solution steps

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

Apply u-substitution

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

= \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 )}

= \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

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

Substitute back

u = e^{x}

= \frac{1}{4} ln ∣ e^{x} ∣ - \frac{1}{4} ln ∣ e^{x} + 4 ∣

\frac{1}{4} ln ∣ e^{x} ∣ = \frac{1}{4} x

= \frac{1}{4} x - \frac{1}{4} ln ∣ e^{x} + 4 ∣

Add a constant to the solution

= \frac{1}{4} x - \frac{1}{4} ln ∣ e^{x} + 4 ∣ + C

x \to 2 lim (x^{3} - 5 x^{2} + 4)

\int \frac{3 x ^{2}}{16 - x ^{2}} d x

y^{^{''}} + 2 y^{^{'}} + y = e^{t}

y^{^{'''}} - 3 y^{^{''}} + y^{^{'}} - 3 y = 0

\int_{- \infty}^{- 1} e^{- 12 t} d t

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