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integral of 1/(x-sqrt(x-2))

Lösung

\int \frac{1}{x - x - 2} d x

ln 4 x - 4 x - 2 + \frac{2}{7} arctan (\frac{2 x - 2 - 1}{7}) + C

Schritte zur Lösung

\int \frac{1}{x - x - 2} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Vervollständige das Quadrat

u^{2} + 2 - u : (u - \frac{1}{2})^{2} + \frac{7}{4}

= 2 \cdot \int \frac{u}{( u - \frac{1}{2} ) ^{2} + \frac{7}{4}} d u

Wende U-Substitution an

= 2 \cdot \int \frac{2 ( 2 v + 1 )}{4 v ^{2} + 7} d v

Entferne die Konstante:

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

= 2 \cdot 2 \cdot \int \frac{2 v + 1}{4 v ^{2} + 7} d v

Multipliziere aus

\frac{2 v + 1}{4 v ^{2} + 7} : \frac{2 v}{4 v ^{2} + 7} + \frac{1}{4 v ^{2} + 7}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 2 (\int \frac{2 v}{4 v ^{2} + 7} d v + \int \frac{1}{4 v ^{2} + 7} d v)

\int \frac{2 v}{4 v ^{2} + 7} d v = \frac{1}{4} ln 4 v^{2} + 7

\int \frac{1}{4 v ^{2} + 7} d v = \frac{1}{2 7} arctan (\frac{2}{7} v)

= 2 \cdot 2 (\frac{1}{4} ln 4 v^{2} + 7 + \frac{1}{2 7} arctan (\frac{2}{7} v))

Ersetze zurück

= 2 \cdot 2 (\frac{1}{4} ln 4 (x - 2 - \frac{1}{2})^{2} + 7 + \frac{1}{2 7} arctan (\frac{2}{7} (x - 2 - \frac{1}{2})))

Vereinfache

2 \cdot 2 (\frac{1}{4} ln 4 (x - 2 - \frac{1}{2})^{2} + 7 + \frac{1}{2 7} arctan (\frac{2}{7} (x - 2 - \frac{1}{2}))) : ln 4 x - 4 x - 2 + \frac{2}{7} arctan (\frac{2 x - 2 - 1}{7})

= ln 4 x - 4 x - 2 + \frac{2}{7} arctan (\frac{2 x - 2 - 1}{7})

Füge eine Konstante zur Lösung hinzu

= ln 4 x - 4 x - 2 + \frac{2}{7} arctan (\frac{2 x - 2 - 1}{7}) + C

\frac{d y}{d x} = \frac{( 3 x + 2 y )}{3 x + 2 y + 2}

\frac{\partial}{\partial x} (x^{4} + x \cdot sin (y))

\int \frac{1}{x ( x - 2 )} d x

\int \frac{6400 q}{16 q ^{2} + 3200} d q

\frac{d}{d x} (\frac{x ^{2} ( x ^{3} - 1 )}{x ^{2} + 1})