integral of x^2(2x-1)^{-7}

Lösung

\int x^{2} (2 x - 1)^{- 7} d x

\frac{- 60 x ^{2} + 12 x - 1}{480 ( 2 x - 1 ) ^{6}} + C

Schritte zur Lösung

\int x^{2} (2 x - 1)^{- 7} d x

Wende U-Substitution an

= \int \frac{( u + 1 ) ^{2}}{8 u ^{7}} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int \frac{( u + 1 ) ^{2}}{u ^{7}} d u

Multipliziere aus

\frac{( u + 1 ) ^{2}}{u ^{7}} : \frac{1}{u ^{5}} + \frac{2}{u ^{6}} + \frac{1}{u ^{7}}

= \frac{1}{8} \cdot \int \frac{1}{u ^{5}} + \frac{2}{u ^{6}} + \frac{1}{u ^{7}} 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}{8} (\int \frac{1}{u ^{5}} d u + \int \frac{2}{u ^{6}} d u + \int \frac{1}{u ^{7}} d u)

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

\int \frac{2}{u ^{6}} d u = - \frac{2}{5 u ^{5}}

\int \frac{1}{u ^{7}} d u = - \frac{1}{6 u ^{6}}

= \frac{1}{8} (- \frac{1}{4 u ^{4}} - \frac{2}{5 u ^{5}} - \frac{1}{6 u ^{6}})

Setze in

u = 2 x - 1

ein

= \frac{1}{8} (- \frac{1}{4 ( 2 x - 1 ) ^{4}} - \frac{2}{5 ( 2 x - 1 ) ^{5}} - \frac{1}{6 ( 2 x - 1 ) ^{6}})

Vereinfache

\frac{1}{8} (- \frac{1}{4 ( 2 x - 1 ) ^{4}} - \frac{2}{5 ( 2 x - 1 ) ^{5}} - \frac{1}{6 ( 2 x - 1 ) ^{6}}) : \frac{- 60 x ^{2} + 12 x - 1}{480 ( 2 x - 1 ) ^{6}}

= \frac{- 60 x ^{2} + 12 x - 1}{480 ( 2 x - 1 ) ^{6}}

Füge eine Konstante zur Lösung hinzu

= \frac{- 60 x ^{2} + 12 x - 1}{480 ( 2 x - 1 ) ^{6}} + C

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