integral from-1 to 1 of ((3e^x))/(1+e^x)

Lösung

\int_{- 1}^{1} \frac{( 3 e ^{x} )}{1 + e ^{x}} d x

3

Schritte zur Lösung

\int_{- 1}^{1} \frac{3 e ^{x}}{1 + e ^{x}} d x

Entferne die Konstante:

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

= 3 \cdot \int_{- 1}^{1} \frac{e ^{x}}{1 + e ^{x}} d x

Wende U-Substitution an

= 3 \cdot \int_{1 + \frac{1}{e}}^{1 + e} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 [ln ∣ u ∣]_{1 + \frac{1}{e}}^{1 + e}

Berechne die Grenzen:

ln (1 + e) - ln (1 + \frac{1}{e})

= 3 (ln (1 + e) - ln (1 + \frac{1}{e}))

Vereinfache

= 3

\int_{0}^{1} 6 x^{5} (9 - x^{6}) d x

\int_{0}^{1} 4 x (x^{2} + 3)^{3} d x

\int_{1}^{2} 2 x ln (x) d x

\int_{0}^{\frac{π}{2}} sin (x) sin (2 x) d x

\int_{0}^{5} (ln (x))^{2} d x