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

Lösung

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

\frac{1}{20} (- ln x^{2} - 2 x + 2 + 4 arctan (x - 1)) + \frac{1}{10} (\frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2})) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{( x ^{2} - 2 x + 2 ) ( x ^{2} + 4 )} : \frac{- x + 3}{10 ( x ^{2} - 2 x + 2 )} + \frac{x - 1}{10 ( x ^{2} + 4 )}

= \int \frac{- x + 3}{10 ( x ^{2} - 2 x + 2 )} + \frac{x - 1}{10 ( x ^{2} + 4 )} 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 + 3}{10 ( x ^{2} - 2 x + 2 )} d x + \int \frac{x - 1}{10 ( x ^{2} + 4 )} d x

\int \frac{- x + 3}{10 ( x ^{2} - 2 x + 2 )} d x = \frac{1}{20} (- ln x^{2} - 2 x + 2 + 4 arctan (x - 1))

\int \frac{x - 1}{10 ( x ^{2} + 4 )} d x = \frac{1}{10} (\frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2}))

= \frac{1}{20} (- ln x^{2} - 2 x + 2 + 4 arctan (x - 1)) + \frac{1}{10} (\frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{20} (- ln x^{2} - 2 x + 2 + 4 arctan (x - 1)) + \frac{1}{10} (\frac{1}{2} ln x^{2} + 4 - \frac{1}{2} arctan (\frac{x}{2})) + C

\int sin (2 x) + 5 tan^{2} (x) csc^{2} (x) d x

\int sec^{2} (x) tan (2 tan (x)) d x

\int 2 x (arcsin (x))^{2} d x

\int - \frac{6}{7 t 49 t ^{2} - 25} d t

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