integral of (x^5-x^4+4x^3-4x^2+8x-4)/((x^2+2)^3)

Lösung

\int \frac{x ^{5} - x ^{4} + 4 x ^{3} - 4 x ^{2} + 8 x - 4}{( x ^{2} + 2 ) ^{3}} d x

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

Schritte zur Lösung

\int \frac{x ^{5} - x ^{4} + 4 x ^{3} - 4 x ^{2} + 8 x - 4}{( x ^{2} + 2 ) ^{3}} d x

Ermittle den Partialbruch von

\frac{x ^{5} - x ^{4} + 4 x ^{3} - 4 x ^{2} + 8 x - 4}{( x ^{2} + 2 ) ^{3}} : \frac{x - 1}{x ^{2} + 2} + \frac{4 x}{( x ^{2} + 2 ) ^{3}}

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e - \frac{6 x}{( x ^{2} - 4 ) ^{2}}

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

\frac{\partial}{\partial x} (x arctan (y))

n = 3 \sum \infty (\frac{9}{11})^{- n}

\int cos^{5} (4 - x) sin (4 - x) d x