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

Lösung

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

- 6 (\frac{( 6 1 - x + 1 ) ^{3}}{3} - \frac{3 ( 6 1 - x + 1 ) ^{2}}{2} + 3 (6 1 - x + 1) - ln 6 1 - x + 1) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int - \frac{1}{u + 3 u} d u

Entferne die Konstante:

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

= - \int \frac{1}{u + 3 u} d u

Wende U-Substitution an

= - \int \frac{6 v ^{3}}{v + 1} d v

Entferne die Konstante:

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

= - 6 \cdot \int \frac{v ^{3}}{v + 1} d v

Wende U-Substitution an

= - 6 \cdot \int \frac{( w - 1 ) ^{3}}{w} d w

Multipliziere aus

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

= - 6 \cdot \int w^{2} - 3 w + 3 - \frac{1}{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

= - 6 (\int w^{2} d w - \int 3 w d w + \int 3 d w - \int \frac{1}{w} d w)

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

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

\int 3 d w = 3 w

\int \frac{1}{w} d w = ln ∣ w ∣

= - 6 (\frac{w ^{3}}{3} - \frac{3 w ^{2}}{2} + 3 w - ln ∣ w ∣)

Ersetze zurück

= - 6 (\frac{( 6 1 - x + 1 ) ^{3}}{3} - \frac{3 ( 6 1 - x + 1 ) ^{2}}{2} + 3 (6 1 - x + 1) - ln 6 1 - x + 1)

Füge eine Konstante zur Lösung hinzu

= - 6 (\frac{( 6 1 - x + 1 ) ^{3}}{3} - \frac{3 ( 6 1 - x + 1 ) ^{2}}{2} + 3 (6 1 - x + 1) - ln 6 1 - x + 1) + C

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

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\int cot^{10} (x) csc^{4} (x) d x

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