integral of (2+w^4)/(w^3+9w)

Lösung

\int \frac{2 + w ^{4}}{w ^{3} + 9 w} d w

\frac{w ^{2}}{2} - \frac{83}{18} ln w^{2} + 9 + \frac{2}{9} ln ∣ w ∣ + C

Schritte zur Lösung

\int \frac{2 + w ^{4}}{w ^{3} + 9 w} d w

Lange Division

\frac{2 + w ^{4}}{w ^{3} + 9 w} : w + \frac{- 9 w ^{2} + 2}{w ^{3} + 9 w}

= \int w + \frac{- 9 w ^{2} + 2}{w ^{3} + 9 w} d w

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int w d w + \int \frac{- 9 w ^{2} + 2}{w ^{3} + 9 w} d w

\int w d w = \frac{w ^{2}}{2}

\int \frac{- 9 w ^{2} + 2}{w ^{3} + 9 w} d w = - \frac{83}{18} ln w^{2} + 9 + \frac{2}{9} ln ∣ w ∣

= \frac{w ^{2}}{2} - \frac{83}{18} ln w^{2} + 9 + \frac{2}{9} ln ∣ w ∣

Füge eine Konstante zur Lösung hinzu

= \frac{w ^{2}}{2} - \frac{83}{18} ln w^{2} + 9 + \frac{2}{9} ln ∣ w ∣ + C

\frac{d y}{d x} + \frac{y}{x} = \frac{1}{x}

d er i v a t i v e sin^{3} (4 x)

y^{^{'}} - 5 y = 0

\frac{d}{d x} (\frac{1}{( 1 - 2500 x ) ^{4}})

n = 1 \sum \infty n^{7} (1 + n^{8})^{0}