integral of (1+2x)^4x

Lösung

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

\frac{x ^{2}}{2} + \frac{8 x ^{3}}{3} + 6 x^{4} + \frac{32 x ^{5}}{5} + \frac{8 x ^{6}}{3} + C

Schritte zur Lösung

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

Multipliziere aus

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

= \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

= \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}

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

Füge eine Konstante zur Lösung hinzu

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

\int (cos (\frac{x}{2}))^{6} d x

in t e g r a l (\frac{ln ( t )}{t})^{2}

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