integral of (e^x)/(sqrt(1-16e^{2x))}

Lösung

\int \frac{e ^{x}}{1 - 16 e ^{2 x}} d x

i \frac{1}{4} ln tan (arcsec (1 - 16 e^{2 x})) + sec (arcsec (1 - 16 e^{2 x})) + C

Schritte zur Lösung

\int \frac{e ^{x}}{1 - 16 e ^{2 x}} d x

Wende U-Substitution an

= \int i \frac{- 1 + u ^{2}}{- 4 + 4 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}}{- 4 + 4 u ^{2}} d u

Vereinfache

\frac{- 1 + u ^{2}}{- 4 + 4 u ^{2}} : \frac{1}{4 - 1 + u ^{2}}

= i \cdot \int \frac{1}{4 - 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}{4} \cdot \int \frac{1}{- 1 + u ^{2}} d u

Trigonometrische Substitution anwenden

= i \frac{1}{4} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= i \frac{1}{4} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= i \frac{1}{4} ln tan (arcsec (1 - 16 e^{2 x})) + sec (arcsec (1 - 16 e^{2 x}))

Füge eine Konstante zur Lösung hinzu

= i \frac{1}{4} ln tan (arcsec (1 - 16 e^{2 x})) + sec (arcsec (1 - 16 e^{2 x})) + C

\int_{9}^{10} \frac{1}{( x - 9 ) ^{\frac{4}{3}}} d x

\int_{0}^{6.25} - x^{2} + 5 x d x

\int 15 x^{\frac{1}{4}} d x

a re a 3 x, \frac{1}{x}, x = 27

\frac{\partial}{\partial y} (y^{2 x})