integral of (x+1)/(sqrt(2x^2-6x+4))

Lösung

\int \frac{x + 1}{2 x ^{2} - 6 x + 4} d x

\frac{1}{2 2} (2 x^{2} - 3 x + 2 + 5 ln 2 x^{2} - 3 x + 2 + 2 x - 3) + C

Schritte zur Lösung

\int \frac{x + 1}{2 x ^{2} - 6 x + 4} d x

Vervollständige das Quadrat

2 x^{2} - 6 x + 4 : 2 (x - \frac{3}{2})^{2} - \frac{1}{2}

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

Wende U-Substitution an

= \int \frac{2 u + 5}{2 4 u ^{2} - 1} 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{2 u + 5}{4 u ^{2} - 1} d u

Multipliziere aus

\frac{2 u + 5}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} + \frac{5}{4 u ^{2} - 1}

= \frac{1}{2} \cdot \int \frac{2 u}{4 u ^{2} - 1} + \frac{5}{4 u ^{2} - 1} d u

Wende die Summenregel an:

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

= \frac{1}{2} (\int \frac{2 u}{4 u ^{2} - 1} d u + \int \frac{5}{4 u ^{2} - 1} d u)

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

\int \frac{5}{4 u ^{2} - 1} d u = \frac{5}{2} ln 4 u^{2} - 1 + 2 u

= \frac{1}{2} (\frac{1}{2} 4 u^{2} - 1 + \frac{5}{2} ln 4 u^{2} - 1 + 2 u)

Setze in

u = x - \frac{3}{2}

ein

= \frac{1}{2} \frac{1}{2} 4 (x - \frac{3}{2})^{2} - 1 + \frac{5}{2} ln 4 (x - \frac{3}{2})^{2} - 1 + 2 (x - \frac{3}{2})

Vereinfache

\frac{1}{2} \frac{1}{2} 4 (x - \frac{3}{2})^{2} - 1 + \frac{5}{2} ln 4 (x - \frac{3}{2})^{2} - 1 + 2 (x - \frac{3}{2}) : \frac{1}{2 2} (2 x^{2} - 3 x + 2 + 5 ln 2 x^{2} - 3 x + 2 + 2 x - 3)

= \frac{1}{2 2} (2 x^{2} - 3 x + 2 + 5 ln 2 x^{2} - 3 x + 2 + 2 x - 3)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2 2} (2 x^{2} - 3 x + 2 + 5 ln 2 x^{2} - 3 x + 2 + 2 x - 3) + C

\frac{d}{d x} (2 π (\frac{x}{9.8})^{\frac{1}{2}})

\int x^{2} + 2 x + 4 d x

d er i v a t i v e x^{4} \cdot cos (4 x)

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

\int \frac{15}{( x + 2 ) ^{4}} d x