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

Lösung

integral

\frac{x ^{3}}{( x ^{2} + x + 2 )}

\frac{x ^{2}}{2} - x - \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{5}{7} arctan (\frac{2 x + 1}{7}) + C

Schritte zur Lösung

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

Lange Division

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

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

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

\int 1 d x = x

\int \frac{- x + 2}{x ^{2} + x + 2} d x = - \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{5}{7} arctan (\frac{2 x + 1}{7})

= \frac{x ^{2}}{2} - x - \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{5}{7} arctan (\frac{2 x + 1}{7})

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{2}}{2} - x - \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{5}{7} arctan (\frac{2 x + 1}{7}) + C

in t e g r a l ln (x - 1)

in t e g r a l sin (\frac{x}{3})

in t e g r a l \frac{1}{( 1 + x ^{2} + y ^{2} )}

in t e g r a l \frac{1}{tan ^{2} ( x )}

in t e g r a l \frac{x ^{3}}{( 1 + x ^{2} ) ^{2}}