What is the integral of 1/(7-2e^x) ?

The integral of 1/(7-2e^x) is -1/7 ln|-7+2e^x|+1/7 (ln(2)+x)+C

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integral of 1/(7-2e^x)

\int \frac{1}{7 - 2 e ^{x}} d x

- \frac{1}{7} ln ∣ - 7 + 2 e^{x} ∣ + \frac{1}{7} (ln (2) + x) + C

Solution steps

\int \frac{1}{7 - 2 e ^{x}} d x

Factor

7 - 2 e^{x} : - (- 7 + 2 e^{x})

= \int \frac{1}{- ( - 7 + 2 e ^{x} )} d x

Take the constant out:

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

= - \int \frac{1}{- 7 + 2 e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( u + 7 )} d u

= - \int \frac{1}{u ( u + 7 )} d u

Take the partial fraction of

\frac{1}{u ( u + 7 )} : \frac{1}{7 u} - \frac{1}{7 ( u + 7 )}

= - \int \frac{1}{7 u} - \frac{1}{7 ( u + 7 )} 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}{7 u} d u - \int \frac{1}{7 ( u + 7 )} d u)

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{1}{7 ( u + 7 )} d u = \frac{1}{7} ln ∣ u + 7 ∣

= - (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

Substitute back

u = - 7 + 2 e^{x}

= - (\frac{1}{7} ln ∣ - 7 + 2 e^{x} ∣ - \frac{1}{7} ln ∣ - 7 + 2 e^{x} + 7 ∣)

Simplify

- (\frac{1}{7} ln ∣ - 7 + 2 e^{x} ∣ - \frac{1}{7} ln ∣ - 7 + 2 e^{x} + 7 ∣) : - \frac{1}{7} ln ∣ - 7 + 2 e^{x} ∣ + \frac{1}{7} (ln (2) + x)

= - \frac{1}{7} ln ∣ - 7 + 2 e^{x} ∣ + \frac{1}{7} (ln (2) + x)

Add a constant to the solution

= - \frac{1}{7} ln ∣ - 7 + 2 e^{x} ∣ + \frac{1}{7} (ln (2) + x) + C

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What is the integral of 1/(7-2e^x) ?
The integral of 1/(7-2e^x) is -1/7 ln|-7+2e^x|+1/7 (ln(2)+x)+C