What is the integral of (35)/(35+e^x) ?

The integral of (35)/(35+e^x) is x-ln|e^x+35|+C

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integral of (35)/(35+e^x)

\int \frac{35}{35 + e ^{x}} d x

x - ln ∣ e^{x} + 35 ∣ + C

Solution steps

\int \frac{35}{35 + e ^{x}} d x

Take the constant out:

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

= 35 \cdot \int \frac{1}{35 + e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( 35 + u )} d u

= 35 \cdot \int \frac{1}{u ( 35 + u )} d u

Take the partial fraction of

\frac{1}{u ( 35 + u )} : \frac{1}{35 u} - \frac{1}{35 ( u + 35 )}

= 35 \cdot \int \frac{1}{35 u} - \frac{1}{35 ( u + 35 )} 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

= 35 (\int \frac{1}{35 u} d u - \int \frac{1}{35 ( u + 35 )} d u)

\int \frac{1}{35 u} d u = \frac{1}{35} ln ∣ u ∣

\int \frac{1}{35 ( u + 35 )} d u = \frac{1}{35} ln ∣ u + 35 ∣

= 35 (\frac{1}{35} ln ∣ u ∣ - \frac{1}{35} ln ∣ u + 35 ∣)

Substitute back

u = e^{x}

= 35 (\frac{1}{35} ln ∣ e^{x} ∣ - \frac{1}{35} ln ∣ e^{x} + 35 ∣)

Simplify

35 (\frac{1}{35} ln ∣ e^{x} ∣ - \frac{1}{35} ln ∣ e^{x} + 35 ∣) : x - ln ∣ e^{x} + 35 ∣

= x - ln ∣ e^{x} + 35 ∣

Add a constant to the solution

= x - ln ∣ e^{x} + 35 ∣ + C

n = 1 \sum \infty 4^{- n}

\int \frac{5 - 50 x}{64 - 25 x ^{2}} d x

x \to 4 + lim (e^{\frac{7}{( 4 - x )}})

4 sin (x) cos (x)

- x^{2} + 7 x, x^{2} - 3 x

What is the integral of (35)/(35+e^x) ?
The integral of (35)/(35+e^x) is x-ln|e^x+35|+C