integral of x(5x+1)^{19}

Lösung

\int x (5 x + 1)^{19} d x

\frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20}) + C

Schritte zur Lösung

\int x (5 x + 1)^{19} d x

Wende U-Substitution an

= \int \frac{u ^{19} ( u - 1 )}{25} d u

Entferne die Konstante:

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

= \frac{1}{25} \cdot \int u^{19} (u - 1) d u

Multipliziere aus

u^{19} (u - 1) : u^{20} - u^{19}

= \frac{1}{25} \cdot \int u^{20} - u^{19} 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}{25} (\int u^{20} d u - \int u^{19} d u)

\int u^{20} d u = \frac{u ^{21}}{21}

\int u^{19} d u = \frac{u ^{20}}{20}

= \frac{1}{25} (\frac{u ^{21}}{21} - \frac{u ^{20}}{20})

Setze in

u = 5 x + 1

ein

= \frac{1}{25} (\frac{( 5 x + 1 ) ^{21}}{21} - \frac{( 5 x + 1 ) ^{20}}{20})

Vereinfache

\frac{1}{25} (\frac{( 5 x + 1 ) ^{21}}{21} - \frac{( 5 x + 1 ) ^{20}}{20}) : \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20})

= \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20}) + C

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