integral from 0 to 4 of 1/(e^x+1)

Lösung

\int_{0}^{4} \frac{1}{e ^{x} + 1} d x

- ln (e^{4} + 1) + ln (2) + 4

Dezimale

0.67499 \dots

Schritte zur Lösung

\int_{0}^{4} \frac{1}{e ^{x} + 1} d x

\frac{1}{e ^{x} + 1} = - \frac{e ^{x}}{e ^{x} + 1} + 1

= \int_{0}^{4} - \frac{e ^{x}}{e ^{x} + 1} + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{4} \frac{e ^{x}}{e ^{x} + 1} d x + \int_{0}^{4} 1 d x

\int_{0}^{4} \frac{e ^{x}}{e ^{x} + 1} d x = ln (e^{4} + 1) - ln (2)

\int_{0}^{4} 1 d x = 4

= - (ln (e^{4} + 1) - ln (2)) + 4

Vereinfache

= - ln (e^{4} + 1) + ln (2) + 4

\int_{0}^{1} \frac{y ^{2}}{2} d y

\int_{2}^{3} \frac{1}{x ^{2} - 6 x + 10} d x

\int_{1}^{4} \frac{x}{1 + x ^{2}} d x

\int_{- 3}^{0} (\frac{y}{3} + y + 4) d y

\int_{0}^{9 π} 3 + sin (x) d x