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

Lösung

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

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

Schritte zur Lösung

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

Lange Division

\frac{2 x ^{3} - x ^{2} + 2 x + 4}{1 + x ^{2}} : 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 (8 x + 3) 4 x^{2} + 3 x d x

\frac{d}{d x} (\frac{x ^{2} + 13 x + 36}{x ^{2} + 8 x - 9})

d er i v a t i v e 9 x^{\frac{2}{3}}

t an g e n t f (x) = 6 sec (x)

\frac{d}{d x} (\frac{ln ( x )}{x ^{4}})