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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{1}{1 + x ^{4}} d x = \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

\int \frac{x ^{2}}{1 + x ^{4}} d x = - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

= \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

Vereinfache

\frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) : \frac{1}{2} arctan (2 x + 1) + \frac{1}{2} arctan (2 x - 1)

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

Füge eine Konstante zur Lösung hinzu

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

t an g e n t f (x) = \frac{1}{x - 1}, at x = - 4

\frac{d}{d x} (e^{x^{3} + 5 x})

\frac{d}{d x} (6^{2 x})

(e^{8 x})^{^{'}}

\frac{\partial}{\partial x} (\frac{x}{x + 3 y})