integral of (e^x)/(sqrt(1-9e^{2x))}

Lösung

\int \frac{e ^{x}}{1 - 9 e ^{2 x}} d x

i \frac{1}{3} ln tan (arcsec (1 - 9 e^{2 x})) + sec (arcsec (1 - 9 e^{2 x})) + C

Schritte zur Lösung

\int \frac{e ^{x}}{1 - 9 e ^{2 x}} d x

Wende U-Substitution an

= \int i \frac{- 1 + u ^{2}}{- 3 + 3 u ^{2}} d u

Entferne die Konstante:

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

= i \cdot \int \frac{- 1 + u ^{2}}{- 3 + 3 u ^{2}} d u

Vereinfache

\frac{- 1 + u ^{2}}{- 3 + 3 u ^{2}} : \frac{1}{3 - 1 + u ^{2}}

= i \cdot \int \frac{1}{3 - 1 + u ^{2}} d u

Entferne die Konstante:

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

= i \frac{1}{3} \cdot \int \frac{1}{- 1 + u ^{2}} d u

Trigonometrische Substitution anwenden

= i \frac{1}{3} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= i \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= i \frac{1}{3} ln tan (arcsec (1 - 9 e^{2 x})) + sec (arcsec (1 - 9 e^{2 x}))

Füge eine Konstante zur Lösung hinzu

= i \frac{1}{3} ln tan (arcsec (1 - 9 e^{2 x})) + sec (arcsec (1 - 9 e^{2 x})) + C

\int \frac{1}{4 x ^{2} + 7} d x

(x, y) \to (0, 0) lim (\frac{2 x}{x ^{2} + x + y ^{2}})

\int \frac{3 x + 5}{x ^{3} - x ^{2} - x + 1} d x

\frac{d}{d x} (arctan (sin (x^{2} + \frac{π}{x})))

y^{^{''}} + y^{^{'}} + y = t