What is the integral of 1/(e^x+3) ?

The integral of 1/(e^x+3) is 1/3 x-1/3 ln|e^x+3|+C

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integral of 1/(e^x+3)

\int \frac{1}{e ^{x} + 3} d x

\frac{1}{3} x - \frac{1}{3} ln ∣ e^{x} + 3 ∣ + C

Solution steps

\int \frac{1}{e ^{x} + 3} d x

Apply u-substitution

: \int \frac{1}{u ( u + 3 )} d u

= \int \frac{1}{u ( u + 3 )} d u

Take the partial fraction of

\frac{1}{u ( u + 3 )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

= \int \frac{1}{3 u} - \frac{1}{3 ( u + 3 )} 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}{3 u} d u - \int \frac{1}{3 ( u + 3 )} d u

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} ln ∣ u + 3 ∣

= \frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣

Substitute back

u = e^{x}

= \frac{1}{3} ln ∣ e^{x} ∣ - \frac{1}{3} ln ∣ e^{x} + 3 ∣

\frac{1}{3} ln ∣ e^{x} ∣ = \frac{1}{3} x

= \frac{1}{3} x - \frac{1}{3} ln ∣ e^{x} + 3 ∣

Add a constant to the solution

= \frac{1}{3} x - \frac{1}{3} ln ∣ e^{x} + 3 ∣ + C

y = sin (x) ln (2 x)

\frac{d}{d x} (\frac{6 arctan ( 2 x )}{5 e ^{3 x}})

t^{3} e^{3 t}

\int ln (\frac{x}{x}) d x

f (x) = a x^{4} + (a - 9) x^{2} - 9

What is the integral of 1/(e^x+3) ?
The integral of 1/(e^x+3) is 1/3 x-1/3 ln|e^x+3|+C