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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

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

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

Wende die Summenregel an:

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

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

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

\int \frac{1}{v} d v = ln ∣ v ∣

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

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

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e ln (x e^{- 3 x})

f (x) = \frac{9}{x}

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

\frac{d}{d x} (\frac{2 x + 6 x ^{2}}{( 1 + 6 x ) ^{2}})

y + x^{2} + y^{2} d x - x d y = 0