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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

t an g e n t 7 x - 4 x

y^{^{'}} - y^{\frac{5}{3}} = 0

\int \frac{x}{x - 20} d x

ma c l a u r in \frac{1}{1 + \frac{1}{1 + x}}

x \to 0 lim (lo g_{2} (x))