integral of 1/(sqrt(2x^2+4x+7))

Lösung

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

\frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{2 x ^{2} + 4 x + 7}{5} + C

Schritte zur Lösung

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

Vervollständige das Quadrat

2 x^{2} + 4 x + 7 : 2 (x + 1)^{2} + 5

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

= \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 (arctan (\frac{2}{5} (x + 1))) + sec (arctan (\frac{2}{5} (x + 1)))

Vereinfache

\frac{1}{2} ln tan (arctan (\frac{2}{5} (x + 1))) + sec (arctan (\frac{2}{5} (x + 1))) : \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{2 x ^{2} + 4 x + 7}{5}

= \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{2 x ^{2} + 4 x + 7}{5}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{2 x ^{2} + 4 x + 7}{5} + C

\frac{1}{6} y^{^{''}} + 6 y = 4 tan (6 t) - \frac{1}{2} e^{3 t}

\frac{d y}{d x} - 1 = y^{2}

\frac{\partial}{\partial x} (x ln (x + 6 y))

y^{^{''}} - 5 y^{^{'}} = e^{5 x}

e x p an d \frac{6}{( w ^{2} - 8 ) ^{4}}