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Beliebt Rechnen >

integral of (1+x)/(1-\sqrt[3]{x)}

Lösung

\int \frac{1 + x}{1 - 3 x} d x

Lösung

- \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} - x - 3 x^{\frac{2}{3}} - 6 3 x - 6 ln - 3 x + 1 + C

Schritte zur Lösung

\int \frac{1 + x}{1 - 3 x} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Lange Division

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

= 3 \cdot \int - u^{4} - u^{3} - u^{2} - 2 u - 2 + \frac{2}{- u + 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

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

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{2} d u = \frac{u ^{3}}{3}

\int 2 u d u = u^{2}

\int 2 d u = 2 u

\int \frac{2}{- u + 1} d u = - 2 ln ∣ - u + 1 ∣

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

Setze in

u = 3 x

ein

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

Vereinfache

3 (- \frac{( 3 x ) ^{5}}{5} - \frac{( 3 x ) ^{4}}{4} - \frac{( 3 x ) ^{3}}{3} - (3 x)^{2} - 2 3 x - 2 ln - 3 x + 1) : - \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} - x - 3 x^{\frac{2}{3}} - 6 3 x - 6 ln - 3 x + 1

= - \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} - x - 3 x^{\frac{2}{3}} - 6 3 x - 6 ln - 3 x + 1

Füge eine Konstante zur Lösung hinzu

= - \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{4}{3}} - x - 3 x^{\frac{2}{3}} - 6 3 x - 6 ln - 3 x + 1 + C

Graph

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