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

Lösung

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

\frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2}) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

\int \frac{1}{3 ( x ^{2} + 4 )} d x = \frac{1}{6} arctan (\frac{x}{2})

= \frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2}) + C

\int \frac{x}{8 x - 7} d x

\frac{\partial}{\partial z} (6 x^{\frac{y}{z}})

in v erse l a pl a ce \frac{1}{s + 3} - \frac{1}{s + 4}

\frac{\partial}{\partial s} (\frac{( 2 r + s )}{t})

\frac{d}{d y} (x sin (x y))