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

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

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

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

2 (2 ln (3) - ln (5))

Decimal

1.17557 \dots

Solution steps

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

Apply u-substitution

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

= 4 \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 )}

= 4 \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

= 4 (\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))

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

Simplify

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

= 2 (2 ln (3) - ln (5))

t \to 0 lim (\frac{4}{t} - \frac{4}{e ^{t} - 1})

x \to 4 lim (2 x - 1)

\frac{d}{d x} (5 e^{4 x})

\int \frac{sin ^{5} ( x )}{x} d x

\int x^{2} 3 7 x^{3} + 1 d x

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