What is the integral from 0 to ln(3) of 3/(e^x+2) ?

The integral from 0 to ln(3) of 3/(e^x+2) is 3*(2ln(3)-ln(5))/2

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integral from 0 to ln(3) of 3/(e^x+2)

\int_{0}^{l n (3)} \frac{3}{e ^{x} + 2} d x

3 \cdot \frac{2 ln ( 3 ) - ln ( 5 )}{2}

Decimal Notation

0.88167 \dots

Solution steps

\int_{0}^{l n (3)} \frac{3}{e ^{x} + 2} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{l n (3)} \frac{1}{e ^{x} + 2} d x

Apply u-substitution

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

= 3 \cdot \int_{1}^{3} \frac{1}{u ( u + 2 )} d u

Take the partial fraction of

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= 3 \cdot \int_{1}^{3} \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} 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

= 3 (\int_{1}^{3} \frac{1}{2 u} d u - \int_{1}^{3} \frac{1}{2 ( u + 2 )} d u)

\int_{1}^{3} \frac{1}{2 u} d u = \frac{1}{2} ln (3)

\int_{1}^{3} \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} (ln (5) - ln (3))

= 3 (\frac{1}{2} ln (3) - \frac{1}{2} (ln (5) - ln (3)))

Join

\frac{1}{2} ln (3) - \frac{1}{2} (ln (5) - ln (3)) : \frac{2 ln ( 3 ) - ln ( 5 )}{2}

= 3 \cdot \frac{2 ln ( 3 ) - ln ( 5 )}{2}

(x - 2) \frac{d y}{d x} - y = 3

y = 9 x - 8 x, (1, 1)

\frac{( 2 - z ) ^{\frac{2}{3}}}{4 z + 5}

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

f (x) = 28 x^{2} - 10 x - 2

What is the integral from 0 to ln(3) of 3/(e^x+2) ?
The integral from 0 to ln(3) of 3/(e^x+2) is 3*(2ln(3)-ln(5))/2