integral of (8x^3-9/x+4e^x)

Lösung

\int (8 x^{3} - \frac{9}{x} + 4 e^{x}) d x

2 x^{4} - 9 ln ∣ x ∣ + 4 e^{x} + C

Schritte zur Lösung

\int 8 x^{3} - \frac{9}{x} + 4 e^{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 8 x^{3} d x - \int \frac{9}{x} d x + \int 4 e^{x} d x

\int 8 x^{3} d x = 2 x^{4}

\int \frac{9}{x} d x = 9 ln ∣ x ∣

\int 4 e^{x} d x = 4 e^{x}

= 2 x^{4} - 9 ln ∣ x ∣ + 4 e^{x}

Füge eine Konstante zur Lösung hinzu

= 2 x^{4} - 9 ln ∣ x ∣ + 4 e^{x} + C

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

\int x csc^{2} (a) x d x

\int \frac{x ^{2} + 1}{x ^{3} + 2 x ^{2} + x} d x

\int \frac{9 x}{( x - 6 ) ( x + 4 )} d x

\int 4 x^{8} (2 x^{9} - 1)^{7} d x