integral of 2x(1+2x)^4

Lösung

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

x^{2} + \frac{16 x ^{3}}{3} + 12 x^{4} + \frac{64 x ^{5}}{5} + \frac{16 x ^{6}}{3} + C

Schritte zur Lösung

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

Entferne die Konstante:

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

= 2 \cdot \int x (1 + 2 x)^{4} d x

Multipliziere aus

x (1 + 2 x)^{4} : x + 8 x^{2} + 24 x^{3} + 32 x^{4} + 16 x^{5}

= 2 \cdot \int x + 8 x^{2} + 24 x^{3} + 32 x^{4} + 16 x^{5} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int x d x + \int 8 x^{2} d x + \int 24 x^{3} d x + \int 32 x^{4} d x + \int 16 x^{5} d x)

\int x d x = \frac{x ^{2}}{2}

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 24 x^{3} d x = 6 x^{4}

\int 32 x^{4} d x = \frac{32 x ^{5}}{5}

\int 16 x^{5} d x = \frac{8 x ^{6}}{3}

= 2 (\frac{x ^{2}}{2} + \frac{8 x ^{3}}{3} + 6 x^{4} + \frac{32 x ^{5}}{5} + \frac{8 x ^{6}}{3})

Vereinfache

2 (\frac{x ^{2}}{2} + \frac{8 x ^{3}}{3} + 6 x^{4} + \frac{32 x ^{5}}{5} + \frac{8 x ^{6}}{3}) : x^{2} + \frac{16 x ^{3}}{3} + 12 x^{4} + \frac{64 x ^{5}}{5} + \frac{16 x ^{6}}{3}

= x^{2} + \frac{16 x ^{3}}{3} + 12 x^{4} + \frac{64 x ^{5}}{5} + \frac{16 x ^{6}}{3}

Füge eine Konstante zur Lösung hinzu

= x^{2} + \frac{16 x ^{3}}{3} + 12 x^{4} + \frac{64 x ^{5}}{5} + \frac{16 x ^{6}}{3} + C

l a pl a ce t r an s f or m 1 - e^{- 2 t}

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