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integral of (x^3)/(x^2-x)

Lösung

\int \frac{x ^{3}}{x ^{2} - x} d x

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

Schritte zur Lösung

\int \frac{x ^{3}}{x ^{2} - x} d x

Vervollständige das Quadrat

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

= \int \frac{x ^{3}}{( x - \frac{1}{2} ) ^{2} - \frac{1}{4}} d x

Wende U-Substitution an

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

Wende U-Substitution an

= \frac{1}{2} \cdot \int \frac{v ^{2}}{2 ( 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 \frac{1}{2} \cdot \int \frac{v ^{2}}{v - 2} d v

Wende U-Substitution an

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{( w + 2 ) ^{2}}{w} d w

Multipliziere aus

\frac{( w + 2 ) ^{2}}{w} : w + 4 + \frac{4}{w}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int w + 4 + \frac{4}{w} d w

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} \cdot \frac{1}{2} (\int w d w + \int 4 d w + \int \frac{4}{w} d w)

\int w d w = \frac{w ^{2}}{2}

\int 4 d w = 4 w

\int \frac{4}{w} d w = 4 ln ∣ w ∣

= \frac{1}{2} \cdot \frac{1}{2} (\frac{w ^{2}}{2} + 4 w + 4 ln ∣ w ∣)

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} ((1 + \frac{1}{x})^{\frac{1}{x}})

\frac{\partial}{\partial x} (x y x^{2} + y^{2})

\int 3 x sin (5 x) d x

a re a x = 9 y^{2}, x = 28 + 2 y^{2}

\int - e^{t} d t