What is the integral of (33)/(33+e^x) ?

The integral of (33)/(33+e^x) is x-ln|e^x+33|+C

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integral of (33)/(33+e^x)

\int \frac{33}{33 + e ^{x}} d x

x - ln ∣ e^{x} + 33 ∣ + C

Solution steps

\int \frac{33}{33 + e ^{x}} d x

Take the constant out:

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

= 33 \cdot \int \frac{1}{33 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 33 + u )} d u

= 33 \cdot \int \frac{1}{u ( 33 + u )} d u

Take the partial fraction of

\frac{1}{u ( 33 + u )} : \frac{1}{33 u} - \frac{1}{33 ( u + 33 )}

= 33 \cdot \int \frac{1}{33 u} - \frac{1}{33 ( u + 33 )} 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

= 33 (\int \frac{1}{33 u} d u - \int \frac{1}{33 ( u + 33 )} d u)

\int \frac{1}{33 u} d u = \frac{1}{33} ln ∣ u ∣

\int \frac{1}{33 ( u + 33 )} d u = \frac{1}{33} ln ∣ u + 33 ∣

= 33 (\frac{1}{33} ln ∣ u ∣ - \frac{1}{33} ln ∣ u + 33 ∣)

Substitute back

u = e^{x}

= 33 (\frac{1}{33} ln ∣ e^{x} ∣ - \frac{1}{33} ln ∣ e^{x} + 33 ∣)

Simplify

33 (\frac{1}{33} ln ∣ e^{x} ∣ - \frac{1}{33} ln ∣ e^{x} + 33 ∣) : x - ln ∣ e^{x} + 33 ∣

= x - ln ∣ e^{x} + 33 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 33 ∣ + C

\int \frac{( x ^{2} - 1 )}{x} d x

\frac{d}{d x} (ln (x^{2} + 1) - e^{\frac{x}{2}} cos (π x))

\int \frac{4 x + 5}{x ^{2} + 4 x + 2} d x

\int \frac{1}{x ^{2} - x} d x

\frac{d y}{d x} = (\frac{( 2 y + 3 )}{( 4 x + 5 )})^{2}

What is the integral of (33)/(33+e^x) ?
The integral of (33)/(33+e^x) is x-ln|e^x+33|+C