integral of 1/(7+e^x)

Lösung

\int \frac{1}{7 + e ^{x}} d x

\frac{1}{7} x - \frac{1}{7} ln ∣ e^{x} + 7 ∣ + C

Schritte zur Lösung

\int \frac{1}{7 + e ^{x}} d x

Wende U-Substitution an

= \int \frac{1}{u ( 7 + u )} d u

Ermittle den Partialbruch von

\frac{1}{u ( 7 + u )} : \frac{1}{7 u} - \frac{1}{7 ( u + 7 )}

= \int \frac{1}{7 u} - \frac{1}{7 ( u + 7 )} d u

Wende die Summenregel an:

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

= \int \frac{1}{7 u} d u - \int \frac{1}{7 ( u + 7 )} d u

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{1}{7 ( u + 7 )} d u = \frac{1}{7} ln ∣ u + 7 ∣

= \frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣

Setze in

u = e^{x}

ein

= \frac{1}{7} ln ∣ e^{x} ∣ - \frac{1}{7} ln ∣ e^{x} + 7 ∣

\frac{1}{7} ln ∣ e^{x} ∣ = \frac{1}{7} x

= \frac{1}{7} x - \frac{1}{7} ln ∣ e^{x} + 7 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{7} x - \frac{1}{7} ln ∣ e^{x} + 7 ∣ + C

\int sec^{2} (5 x) tan^{2} (5 x) d x

\int \frac{1}{( x ^{2} - 6 x + 10 ) ^{5}} d x

\int \frac{x ^{2}}{( 2 - 2 x ^{3} ) ^{3}} d x

\int e^{9 t} d t

\int x^{3} lo g_{10} (x) d x