What is the integral from 0 to 3 of 1/(3+2^x) ?

The integral from 0 to 3 of 1/(3+2^x) is (5ln(2)-ln(11))/(3ln(2))

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integral from 0 to 3 of 1/(3+2^x)

\int_{0}^{3} \frac{1}{3 + 2 ^{x}} d x

\frac{5 ln ( 2 ) - ln ( 11 )}{3 ln ( 2 )}

Decimal

0.51352 \dots

Solution steps

\int_{0}^{3} \frac{1}{3 + 2 ^{x}} d x

Apply u-substitution

= \int_{1}^{8} \frac{1}{ln ( 2 ) u ( 3 + u )} d u

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int_{1}^{8} \frac{1}{u ( 3 + u )} d u

Take the partial fraction of

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

= \frac{1}{ln ( 2 )} \cdot \int_{1}^{8} \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

= \frac{1}{ln ( 2 )} (\int_{1}^{8} \frac{1}{3 u} d u - \int_{1}^{8} \frac{1}{3 ( u + 3 )} d u)

\int_{1}^{8} \frac{1}{3 u} d u = ln (2)

\int_{1}^{8} \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} (ln (11) - 2 ln (2))

= \frac{1}{ln ( 2 )} (ln (2) - \frac{1}{3} (ln (11) - 2 ln (2)))

Simplify

\frac{1}{ln ( 2 )} (ln (2) - \frac{1}{3} (ln (11) - 2 ln (2))) : \frac{5 ln ( 2 ) - ln ( 11 )}{3 ln ( 2 )}

= \frac{5 ln ( 2 ) - ln ( 11 )}{3 ln ( 2 )}

What is the integral from 0 to 3 of 1/(3+2^x) ?
The integral from 0 to 3 of 1/(3+2^x) is (5ln(2)-ln(11))/(3ln(2))