integral of (x^2+2x+3)/((x-1)(x+1)^2)

Lösung

\int \frac{x ^{2} + 2 x + 3}{( x - 1 ) ( x + 1 ) ^{2}} d x

\frac{3}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + C

Schritte zur Lösung

\int \frac{x ^{2} + 2 x + 3}{( x - 1 ) ( x + 1 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{x ^{2} + 2 x + 3}{( x - 1 ) ( x + 1 ) ^{2}} : \frac{3}{2 ( x - 1 )} - \frac{1}{2 ( x + 1 )} - \frac{1}{( x + 1 ) ^{2}}

= \int \frac{3}{2 ( x - 1 )} - \frac{1}{2 ( x + 1 )} - \frac{1}{( x + 1 ) ^{2}} 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{3}{2 ( x - 1 )} d x - \int \frac{1}{2 ( x + 1 )} d x - \int \frac{1}{( x + 1 ) ^{2}} d x

\int \frac{3}{2 ( x - 1 )} d x = \frac{3}{2} ln ∣ x - 1 ∣

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

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

= \frac{3}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ - (- \frac{1}{x + 1})

Vereinfache

= \frac{3}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{x + 1}

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} ln ∣ x - 1 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + C

x \to 0 lim ((sin (x))^{2})

\frac{d}{d x} (3 + \frac{2}{3} (1 - x)^{\frac{3}{2}})

\int \frac{sin ( \frac{1}{x ^{7}} )}{x ^{8}} d x

\int \frac{sin ^{7} ( 7 x )}{cos ^{4} ( 7 x )} d x

d er i v a t i v e ln (12 x)