integral of (2x^3-x^2+2x+4)/(x^2+1)

Lösung

\int \frac{2 x ^{3} - x ^{2} + 2 x + 4}{x ^{2} + 1} d x

x^{2} - x + 5 arctan (x) + C

Schritte zur Lösung

\int \frac{2 x ^{3} - x ^{2} + 2 x + 4}{x ^{2} + 1} d x

Lange Division

\frac{2 x ^{3} - x ^{2} + 2 x + 4}{x ^{2} + 1} : 2 x - 1 + \frac{5}{x ^{2} + 1}

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

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

\int 1 d x = x

\int \frac{5}{x ^{2} + 1} d x = 5 arctan (x)

= x^{2} - x + 5 arctan (x)

Füge eine Konstante zur Lösung hinzu

= x^{2} - x + 5 arctan (x) + C

\int e^{20 x} d x

t an g e n t f (x) = 2 x^{4}, at x = - 2

\int cos (x) ln (x) d x

\frac{\partial}{\partial x} (x cos (y) + y)

\frac{d}{d x} (e^{\frac{x}{8}} cos (\frac{x}{8}))