integral (x+4x-5i)^3

Lösung

integral

(x + 4 x - 5 i)^{3}

\frac{1}{4} (125 x^{4} - 750 x^{2}) + i (125 x - 125 x^{3}) + C

Schritte zur Lösung

\int (x + 4 x - 5 i)^{3} d x

Multipliziere aus

(x + 4 x - 5 i)^{3} : (125 x^{3} - 375 x) + i (125 - 375 x^{2})

= \int 125 x^{3} - 375 x + i (125 - 375 x^{2}) 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 125 x^{3} d x - \int 375 x d x + \int i (125 - 375 x^{2}) d x

\int 125 x^{3} d x = \frac{125 x ^{4}}{4}

\int 375 x d x = \frac{375 x ^{2}}{2}

\int i (125 - 375 x^{2}) d x = i (125 x - 125 x^{3})

= \frac{125 x ^{4}}{4} - \frac{375 x ^{2}}{2} + i (125 x - 125 x^{3})

Vereinfache

\frac{125 x ^{4}}{4} - \frac{375 x ^{2}}{2} + i (125 x - 125 x^{3}) : \frac{1}{4} (125 x^{4} - 750 x^{2}) + i (125 x - 125 x^{3})

= \frac{1}{4} (125 x^{4} - 750 x^{2}) + i (125 x - 125 x^{3})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (125 x^{4} - 750 x^{2}) + i (125 x - 125 x^{3}) + C

in t e g r a l \frac{1}{( 2 - y ^{2} )}

in t e g r a l \frac{1}{( 1 + sin ( x ) )}

in t e g r a l x^{- 1} 0.01

in t e g r a l \frac{x}{( x + b ) ^{3}}

in t e g r a l y = 2 e^{4 x - 3} 6