integral of 1/(sqrt(5x^2+20x-10))

Lösung

\int \frac{1}{5 x ^{2} + 20 x - 10} d x

\frac{1}{5} ln \frac{x ^{2} + 4 x - 2}{6} + \frac{x + 2}{6} + C

Schritte zur Lösung

\int \frac{1}{5 x ^{2} + 20 x - 10} d x

Vervollständige das Quadrat

5 x^{2} + 20 x - 10 : 5 (x + 2)^{2} - 30

= \int \frac{1}{5 ( x + 2 ) ^{2} - 30} d x

Wende U-Substitution an

= \int \frac{1}{5 u ^{2} - 30} d u

Trigonometrische Substitution anwenden

= \int \frac{sec ( v )}{5} d v

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= \frac{1}{5} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= \frac{1}{5} ln tan (arcsec (\frac{5}{30} (x + 2))) + sec (arcsec (\frac{5}{30} (x + 2)))

Vereinfache

\frac{1}{5} ln tan (arcsec (\frac{5}{30} (x + 2))) + sec (arcsec (\frac{5}{30} (x + 2))) : \frac{1}{5} ln \frac{x ^{2} + 4 x - 2}{6} + \frac{x + 2}{6}

= \frac{1}{5} ln \frac{x ^{2} + 4 x - 2}{6} + \frac{x + 2}{6}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} ln \frac{x ^{2} + 4 x - 2}{6} + \frac{x + 2}{6} + C

x \to \frac{π}{3} lim (\frac{1}{sec ( x )})

\frac{d y}{cos ( x ) + y} = d x

\int_{0}^{1} sin^{2} (π x) d x

\frac{\partial}{\partial x} (arctan (x + 2 y))

\int \frac{1}{x} ln (x^{2}) d x