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

Lösung

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

- \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{1}{4 x ^{2}} d x = - \frac{1}{4 x}

\int \frac{1}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} arctan (\frac{x}{2})

= - \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}) + C

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{- t} cos (2 t)

x \cdot d x - y^{2} \cdot d y = 0

y^{^{'}} = x^{2} + 1

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{15}{2}}} d t

\int (x^{6} - 2 x^{3})^{5} (x^{5} - x^{2}) d x