integral of 1/(sqrt(24x+6x^2))

Lösung

\int \frac{1}{24 x + 6 x ^{2}} d x

\frac{1}{6} ln (\frac{x ^{2} + 4 x + x + 2}{2}) + C

Schritte zur Lösung

\int \frac{1}{24 x + 6 x ^{2}} d x

Vervollständige das Quadrat

24 x + 6 x^{2} : 6 (x + 2)^{2} - 24

= \int \frac{1}{6 ( x + 2 ) ^{2} - 24} d x

Wende U-Substitution an

= \int \frac{1}{6 u ^{2} - 24} d u

Trigonometrische Substitution anwenden

= \int \frac{sec ( v )}{6} d v

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= \frac{1}{6} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= \frac{1}{6} ln tan (arcsec (\frac{6}{2 6} (x + 2))) + sec (arcsec (\frac{6}{2 6} (x + 2)))

Vereinfache

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

= \frac{1}{6} ln (\frac{x ^{2} + 4 x + x + 2}{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln (\frac{x ^{2} + 4 x + x + 2}{2}) + C

\int \frac{7}{7 x ^{2} + 14 x + 5} d x

x \to 0 lim (\frac{1}{x})

\frac{\partial}{\partial y} (e^{y} \cdot cos (x))

in v erse l a pl a ce \frac{( s + 3 )}{s ^{2} + 2 s + 1}

4 y^{^{''}} - 16 y^{^{'}} + 18 y = 0