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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

= - \frac{1}{x} - arctan (x)

Füge eine Konstante zur Lösung hinzu

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

\frac{\partial}{\partial x} (t^{2} x - t)

\int \frac{x ^{2}}{x ^{4} - 20 x ^{2} - 125} d x

\frac{d}{d x} (5 \cdot 4^{x})

\int e^{(a - b) x} d x

\int \frac{2 ln ^{2} ( x ) + 16}{x} d x