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

Lösung

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

Füge eine Konstante zur Lösung hinzu

= \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)) + C

\int \frac{5}{x ^{2} + 1} d x

\int sec (\frac{x}{2}) d x

d er i v a t i v e 7 e^{x} + \frac{4}{3 x}

t an g e n t f (x) = 110 x - 16 x^{2}, at 3.438

d er i v a t i v e \frac{x ^{4} - 16 x ^{2} - 16}{( x ^{2} - 4 ) ^{2}}