integral of (4-3x)^5+e^{2x+5}+sin(9x)

Lösung

\int (4 - 3 x)^{5} + e^{2 x + 5} + sin (9 x) d x

- \frac{1}{18} (4 - 3 x)^{6} + \frac{1}{2} e^{2 x + 5} - \frac{1}{9} cos (9 x) + C

Schritte zur Lösung

\int (4 - 3 x)^{5} + e^{2 x + 5} + sin (9 x) 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 (4 - 3 x)^{5} d x + \int e^{2 x + 5} d x + \int sin (9 x) d x

\int (4 - 3 x)^{5} d x = - \frac{1}{18} (4 - 3 x)^{6}

\int e^{2 x + 5} d x = \frac{1}{2} e^{2 x + 5}

\int sin (9 x) d x = - \frac{1}{9} cos (9 x)

= - \frac{1}{18} (4 - 3 x)^{6} + \frac{1}{2} e^{2 x + 5} - \frac{1}{9} cos (9 x)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{18} (4 - 3 x)^{6} + \frac{1}{2} e^{2 x + 5} - \frac{1}{9} cos (9 x) + C

y^{^{'}} + \frac{y}{x} = x

\int_{1}^{\infty} \frac{3}{x ^{5}} d x

\frac{d}{d y} (2 x y^{2} - 3)

d er i v a t i v e t^{4}

x \to 0 + lim (\frac{1}{x ^{2}} - \frac{1}{x})