integral of (x^3+1)/((x^2-4x+5)^2)

Lösung

\int \frac{x ^{3} + 1}{( x ^{2} - 4 x + 5 ) ^{2}} d x

\frac{1}{2} ln x^{2} - 4 x + 5 + \frac{15}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5} + C

Schritte zur Lösung

\int \frac{x ^{3} + 1}{( x ^{2} - 4 x + 5 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{x ^{3} + 1}{( x ^{2} - 4 x + 5 ) ^{2}} : \frac{x + 4}{x ^{2} - 4 x + 5} + \frac{11 x - 19}{( x ^{2} - 4 x + 5 ) ^{2}}

= \int \frac{x + 4}{x ^{2} - 4 x + 5} + \frac{11 x - 19}{( x ^{2} - 4 x + 5 ) ^{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{x + 4}{x ^{2} - 4 x + 5} d x + \int \frac{11 x - 19}{( x ^{2} - 4 x + 5 ) ^{2}} d x

\int \frac{x + 4}{x ^{2} - 4 x + 5} d x = \frac{1}{2} ln x^{2} - 4 x + 5 + 6 arctan (x - 2)

\int \frac{11 x - 19}{( x ^{2} - 4 x + 5 ) ^{2}} d x = \frac{3}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5}

= \frac{1}{2} ln x^{2} - 4 x + 5 + 6 arctan (x - 2) + \frac{3}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5}

Addiere gleiche Elemente:

6 arctan (x - 2) + \frac{3}{2} arctan (x - 2) = \frac{15}{2} arctan (x - 2)

= \frac{1}{2} ln x^{2} - 4 x + 5 + \frac{15}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln x^{2} - 4 x + 5 + \frac{15}{2} arctan (x - 2) - \frac{19}{4} sin (2 arctan (x - 2)) - \frac{11}{2 ( x ^{2} - 4 x + 5 )} + \frac{11 ( x - 2 )}{x ^{2} - 4 x + 5} + C

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

\int \frac{2 x ^{2}}{x ^{3} x ^{6} - 4} d x

\int \frac{5 x ^{2} - 36 x + 18}{x ( x - 4 ) ^{2}} d x

\int \frac{1}{6 x ^{3} - 7 x ^{2} + 2 x} d x

\int \frac{x - 4}{x ( x - 2 ) ( x - 5 )} d x