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

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Wende U-Substitution an

= 2 \cdot \int \frac{1}{( u - 1 ) u ^{2}} d u

Ermittle den Partialbruch von

\frac{1}{( u - 1 ) u ^{2}} : \frac{1}{u - 1} - \frac{1}{u} - \frac{1}{u ^{2}}

= 2 \cdot \int \frac{1}{u - 1} - \frac{1}{u} - \frac{1}{u ^{2}} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 2 (ln ∣ u - 1 ∣ - ln ∣ u ∣ - (- \frac{1}{u}))

Setze in

u = x + 1

ein

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

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