integral of (x^2)/((1-x)^{2021)}

Lösung

\int \frac{x ^{2}}{( 1 - x ) ^{2021}} d x

\frac{1}{2018 ( 1 - x ) ^{2018}} - \frac{2}{2019 ( 1 - x ) ^{2019}} + \frac{1}{2020 ( 1 - x ) ^{2020}} + C

Schritte zur Lösung

\int \frac{x ^{2}}{( 1 - x ) ^{2021}} d x

Wende U-Substitution an

= \int - \frac{( - u + 1 ) ^{2}}{u ^{2021}} d u

Entferne die Konstante:

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

= - \int \frac{( - u + 1 ) ^{2}}{u ^{2021}} d u

Multipliziere aus

\frac{( - u + 1 ) ^{2}}{u ^{2021}} : \frac{1}{u ^{2019}} - \frac{2}{u ^{2020}} + \frac{1}{u ^{2021}}

= - \int \frac{1}{u ^{2019}} - \frac{2}{u ^{2020}} + \frac{1}{u ^{2021}} 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

= - (\int \frac{1}{u ^{2019}} d u - \int \frac{2}{u ^{2020}} d u + \int \frac{1}{u ^{2021}} d u)

\int \frac{1}{u ^{2019}} d u = - \frac{1}{2018 u ^{2018}}

\int \frac{2}{u ^{2020}} d u = - \frac{2}{2019 u ^{2019}}

\int \frac{1}{u ^{2021}} d u = - \frac{1}{2020 u ^{2020}}

= - (- \frac{1}{2018 u ^{2018}} - (- \frac{2}{2019 u ^{2019}}) - \frac{1}{2020 u ^{2020}})

Setze in

u = 1 - x

ein

= - (- \frac{1}{2018 ( 1 - x ) ^{2018}} - (- \frac{2}{2019 ( 1 - x ) ^{2019}}) - \frac{1}{2020 ( 1 - x ) ^{2020}})

Vereinfache

= \frac{1}{2018 ( 1 - x ) ^{2018}} - \frac{2}{2019 ( 1 - x ) ^{2019}} + \frac{1}{2020 ( 1 - x ) ^{2020}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2018 ( 1 - x ) ^{2018}} - \frac{2}{2019 ( 1 - x ) ^{2019}} + \frac{1}{2020 ( 1 - x ) ^{2020}} + C

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