What is the integral of 6/(2e^x+5) ?

The integral of 6/(2e^x+5) is -6/5 (-x+ln| 4/5 e^x+2|-ln(4/5))+C

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integral of 6/(2e^x+5)

\int \frac{6}{2 e ^{x} + 5} d x

- \frac{6}{5} (- x + ln \frac{4}{5} e^{x} + 2 - ln (\frac{4}{5})) + C

Solution steps

\int \frac{6}{2 e ^{x} + 5} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{2 e ^{x} + 5} d x

Apply u-substitution

= 6 \cdot \int \frac{1}{u ( 2 u + 5 )} d u

Complete the square

u (2 u + 5) : 2 (u + \frac{5}{4})^{2} - \frac{25}{8}

= 6 \cdot \int \frac{1}{2 ( u + \frac{5}{4} ) ^{2} - \frac{25}{8}} d u

Apply u-substitution

= 6 \cdot \int \frac{8}{16 v ^{2} - 25} d v

Take the constant out:

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

= 6 \cdot 8 \cdot \int \frac{1}{16 v ^{2} - 25} d v

Apply Integral Substitution

= 6 \cdot 8 \cdot \int \frac{1}{20 ( w ^{2} - 1 )} d w

Take the constant out:

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

= 6 \cdot 8 \cdot \frac{1}{20} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

w^{2} - 1 : - (- w^{2} + 1)

= 6 \cdot 8 \cdot \frac{1}{20} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

Take the constant out:

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

= 6 \cdot 8 \cdot \frac{1}{20} (- \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 6 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 6 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln \frac{4}{5} ( e ^{x} + \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( e ^{x} + \frac{5}{4} ) - 1}{2}))

Simplify

6 \cdot 8 \cdot \frac{1}{20} (- (\frac{ln \frac{4}{5} ( e ^{x} + \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( e ^{x} + \frac{5}{4} ) - 1}{2})) : - \frac{6}{5} (- x + ln \frac{4}{5} e^{x} + 2 - ln (\frac{4}{5}))

= - \frac{6}{5} (- x + ln \frac{4}{5} e^{x} + 2 - ln (\frac{4}{5}))

Add a constant to the solution

= - \frac{6}{5} (- x + ln \frac{4}{5} e^{x} + 2 - ln (\frac{4}{5})) + C

What is the integral of 6/(2e^x+5) ?
The integral of 6/(2e^x+5) is -6/5 (-x+ln| 4/5 e^x+2|-ln(4/5))+C