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

Lösung

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

- \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + C

\int \frac{t ^{2}}{3 6 + t ^{3}} d t

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\int \frac{x ^{2} + 4 x}{( x + 2 ) ^{2}} d x

\int 2 x^{2} - 8 d x

\int (x^{2} + 3 x) (sin (x^{2})) d x