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

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

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

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

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

Decimal Notation

0.62667 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{u ( u + 3 )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

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

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

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

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

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

Join

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

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

\frac{d y}{d x} x^{2} + y x = 1, y (6) = 9

\int e^{e} d e

x (\frac{d y}{d x}) = 1 - y^{2}

x \to \infty lim ((\frac{x + 1}{x + 3})^{\frac{x}{5}})

\frac{d y}{d x} + 9 y = e^{x} y^{- 4}

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