What is the integral of 1/(2e^x+3) ?

The integral of 1/(2e^x+3) is -1/3 (ln| 4/3 e^x+2|-ln(4/3)-x)+C

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integral of 1/(2e^x+3)

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

- \frac{1}{3} (ln \frac{4}{3} e^{x} + 2 - ln (\frac{4}{3}) - x) + C

Solution steps

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

Apply u-substitution

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

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

Complete the square

u (2 u + 3) : 2 (u + \frac{3}{4})^{2} - \frac{9}{8}

= \int \frac{1}{2 ( u + \frac{3}{4} ) ^{2} - \frac{9}{8}} d u

Apply u-substitution

: \int \frac{8}{16 v ^{2} - 9} d v

= \int \frac{8}{16 v ^{2} - 9} d v

Take the constant out:

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

= 8 \cdot \int \frac{1}{16 v ^{2} - 9} d v

Apply Integral Substitution

= 8 \cdot \int \frac{1}{12 ( w ^{2} - 1 )} d w

Take the constant out:

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

= 8 \cdot \frac{1}{12} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 8 \cdot \frac{1}{12} \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

= 8 \cdot \frac{1}{12} (- \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}

= 8 \cdot \frac{1}{12} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 8 \cdot \frac{1}{12} (- (\frac{ln \frac{4}{3} ( e ^{x} + \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{3} ( e ^{x} + \frac{3}{4} ) - 1}{2}))

Simplify

8 \cdot \frac{1}{12} (- (\frac{ln \frac{4}{3} ( e ^{x} + \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{3} ( e ^{x} + \frac{3}{4} ) - 1}{2})) : - \frac{1}{3} (ln \frac{4}{3} e^{x} + 2 - ln (\frac{4}{3}) - x)

= - \frac{1}{3} (ln \frac{4}{3} e^{x} + 2 - ln (\frac{4}{3}) - x)

Add a constant to the solution

= - \frac{1}{3} (ln \frac{4}{3} e^{x} + 2 - ln (\frac{4}{3}) - x) + C

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What is the integral of 1/(2e^x+3) ?
The integral of 1/(2e^x+3) is -1/3 (ln| 4/3 e^x+2|-ln(4/3)-x)+C