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

Lösung

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

- \frac{2}{5} ln ∣ x + 2 ∣ + \frac{1}{5} (ln x^{2} + 1 + arctan (x)) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{x}{( x + 2 ) ( x ^{2} + 1 )} : - \frac{2}{5 ( x + 2 )} + \frac{2 x + 1}{5 ( x ^{2} + 1 )}

= \int - \frac{2}{5 ( x + 2 )} + \frac{2 x + 1}{5 ( 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{2}{5 ( x + 2 )} d x + \int \frac{2 x + 1}{5 ( x ^{2} + 1 )} d x

\int \frac{2}{5 ( x + 2 )} d x = \frac{2}{5} ln ∣ x + 2 ∣

\int \frac{2 x + 1}{5 ( x ^{2} + 1 )} d x = \frac{1}{5} (ln x^{2} + 1 + arctan (x))

= - \frac{2}{5} ln ∣ x + 2 ∣ + \frac{1}{5} (ln x^{2} + 1 + arctan (x))

Füge eine Konstante zur Lösung hinzu

= - \frac{2}{5} ln ∣ x + 2 ∣ + \frac{1}{5} (ln x^{2} + 1 + arctan (x)) + C

d er i v a t i v e y = f (x) = cos (3 x^{2})

6 y^{^{''}} - y^{^{'}} - y = 0, y (0) = 10, y^{^{'}} (0) = 0

x \to 2.1 lim (- x^{2} + 5 x - 2)

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

\frac{d}{d x} (\frac{1}{a - x})