integral of 1/(sqrt(8x+2x^2))

Lösung

\int \frac{1}{8 x + 2 x ^{2}} d x

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

Schritte zur Lösung

\int \frac{1}{8 x + 2 x ^{2}} d x

Vervollständige das Quadrat

8 x + 2 x^{2} : 2 (x + 2)^{2} - 8

= \int \frac{1}{2 ( x + 2 ) ^{2} - 8} d x

Wende U-Substitution an

= \int \frac{1}{2 u ^{2} - 8} d u

\frac{1}{2 u ^{2} - 8} = \frac{1}{2} \frac{1}{u ^{2} - 4}

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

Entferne die Konstante:

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

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

Trigonometrische Substitution anwenden

= \frac{1}{2} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

x \to 2 lim (2 x^{2} + \frac{a}{2} x - 4)

d er i v a t i v e (x^{2} - x + 2)^{3}

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

l a pl a ce t r an s f or m f (t) = t^{2} e^{, at}

y^{^{'}} + y x^{3} = 4 x^{3}