integral of 1/(2x^3+x)

Lösung

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

ln ∣ x ∣ - \frac{1}{2} ln 2 x^{2} + 1 + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{2 x ^{3} + x} : \frac{1}{x} - \frac{2 x}{2 x ^{2} + 1}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{1}{2} ln 2 x^{2} + 1

Füge eine Konstante zur Lösung hinzu

= ln ∣ x ∣ - \frac{1}{2} ln 2 x^{2} + 1 + C

\frac{\partial}{\partial x} ((0.5 x^{2} + 0.5 y^{2})^{\frac{1}{2}})

\int_{0}^{\frac{π}{2}} cos^{5} (1) d x

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 6} e^{2 t}

\frac{\partial}{\partial y} (x e^{x^{2} y})

y^{^{''}} - 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 5