What is the integral from 0 to ln(2) of 3/(e^x+4) ?

The integral from 0 to ln(2) of 3/(e^x+4) is 3*(ln(2)-ln(6)+ln(5))/4

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integral from 0 to ln(2) of 3/(e^x+4)

\int_{0}^{l n (2)} \frac{3}{e ^{x} + 4} d x

3 \cdot \frac{ln ( 2 ) - ln ( 6 ) + ln ( 5 )}{4}

Decimal Notation

0.38311 \dots

Solution steps

\int_{0}^{l n (2)} \frac{3}{e ^{x} + 4} 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 (2)} \frac{1}{e ^{x} + 4} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{u ( u + 4 )} d u

= 3 \cdot \int_{1}^{2} \frac{1}{u ( u + 4 )} d u

Take the partial fraction of

\frac{1}{u ( u + 4 )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

= 3 \cdot \int_{1}^{2} \frac{1}{4 u} - \frac{1}{4 ( u + 4 )} 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}^{2} \frac{1}{4 u} d u - \int_{1}^{2} \frac{1}{4 ( u + 4 )} d u)

\int_{1}^{2} \frac{1}{4 u} d u = \frac{1}{4} ln (2)

\int_{1}^{2} \frac{1}{4 ( u + 4 )} d u = \frac{1}{4} (ln (6) - ln (5))

= 3 (\frac{1}{4} ln (2) - \frac{1}{4} (ln (6) - ln (5)))

Join

\frac{1}{4} ln (2) - \frac{1}{4} (ln (6) - ln (5)) : \frac{ln ( 2 ) - ln ( 6 ) + ln ( 5 )}{4}

= 3 \cdot \frac{ln ( 2 ) - ln ( 6 ) + ln ( 5 )}{4}

f (x) = x^{2} - 4 x - 4

\frac{d}{d x} (0^{4})

\frac{d}{d x} ((5 x^{3} - 1)^{3} ln (x^{2} + 1))

\int 3 x sin (x) d x

x \to 0 lim (x^{6} cos (\frac{8}{x}))

What is the integral from 0 to ln(2) of 3/(e^x+4) ?
The integral from 0 to ln(2) of 3/(e^x+4) is 3*(ln(2)-ln(6)+ln(5))/4