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

Lösung

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

arctan (x) + \frac{1}{2} ln x^{2} + 2 + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

= arctan (x) + \frac{1}{2} ln x^{2} + 2

Füge eine Konstante zur Lösung hinzu

= arctan (x) + \frac{1}{2} ln x^{2} + 2 + C

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

d er i v a t i v e \frac{5}{( 2 x ) ^{3}}

\int_{0}^{\frac{1}{4}} x arcsin (4 x) d x

d er i v a t i v e \frac{( m ^{2} )}{4} + \frac{( 3 m )}{2} + 2

d y - (y - 1) d x = 0