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

Lösung

\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)) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{x}{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 ( 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{1}{1 - 9 x ^{2}} d x

\frac{d y}{d x} = e^{3 x + 2 y}

n \to \infty lim (\frac{n}{6} sin (\frac{6}{n}))

\int e^{t a n (5 x)} sec^{2} (5 x) d x

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