What is the integral of 2/((3+e^x)) ?

The integral of 2/((3+e^x)) is 2(1/3 x-1/3 ln|e^x+3|)+C

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integral of 2/((3+e^x))

integral

\frac{2}{( 3 + e ^{x} )}

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

Solution steps

\int \frac{2}{3 + e ^{x}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{3 + e ^{x}} d x

Apply u-substitution

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

= 2 \cdot \int \frac{1}{u ( 3 + u )} d u

Take the partial fraction of

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

= 2 \cdot \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

= 2 (\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 ∣

= 2 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣)

Substitute back

u = e^{x}

= 2 (\frac{1}{3} ln ∣ e^{x} ∣ - \frac{1}{3} ln ∣ e^{x} + 3 ∣)

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

= 2 (\frac{1}{3} x - \frac{1}{3} ln ∣ e^{x} + 3 ∣)

Add a constant to the solution

= 2 (\frac{1}{3} x - \frac{1}{3} ln ∣ e^{x} + 3 ∣) + C

\frac{sin ( y )}{cos ^{2} ( y )}

tan (x) + \frac{1}{x}

e^{\frac{- y}{2}}

\frac{e ^{2} x}{2}

1 - tan^{2} (x)

What is the integral of 2/((3+e^x)) ?
The integral of 2/((3+e^x)) is 2(1/3 x-1/3 ln|e^x+3|)+C