integral of (4x^5-2x+3)/(x^3)

Lösung

\int \frac{4 x ^{5} - 2 x + 3}{x ^{3}} d x

\frac{4 x ^{3}}{3} + \frac{2}{x} - \frac{3}{2 x ^{2}} + C

Schritte zur Lösung

\int \frac{4 x ^{5} - 2 x + 3}{x ^{3}} d x

Multipliziere aus

\frac{4 x ^{5} - 2 x + 3}{x ^{3}} : 4 x^{2} - \frac{2}{x ^{2}} + \frac{3}{x ^{3}}

= \int 4 x^{2} - \frac{2}{x ^{2}} + \frac{3}{x ^{3}} 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 x^{2} d x - \int \frac{2}{x ^{2}} d x + \int \frac{3}{x ^{3}} d x

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

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

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

= \frac{4 x ^{3}}{3} - (- \frac{2}{x}) - \frac{3}{2 x ^{2}}

Vereinfache

= \frac{4 x ^{3}}{3} + \frac{2}{x} - \frac{3}{2 x ^{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{4 x ^{3}}{3} + \frac{2}{x} - \frac{3}{2 x ^{2}} + C

in v erse l a pl a ce e^{- 2 s} (\frac{1}{s - 9})

d er i v a t i v e y = (2 x + 1)^{\frac{2}{3}} (3 x - 4)^{\frac{1}{2}}

\frac{d}{d t} (e^{t} \cdot cos (t))

s l o p e (\frac{1}{2}, 2), (6, 2)

\int_{- 3}^{0} \frac{1}{9 + 2 x} d x