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

Lösung

integral

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

2 (\frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{3 ( x + 1 )} + \frac{1}{9} (- \frac{1}{2} ln x^{2} + 2 + 22 arctan (\frac{x}{2}))) + C

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

\frac{x}{( x + 1 ) ^{2} ( x ^{2} + 2 )} : \frac{1}{9 ( x + 1 )} - \frac{1}{3 ( x + 1 ) ^{2}} + \frac{- x + 4}{9 ( x ^{2} + 2 )}

= 2 \cdot \int \frac{1}{9 ( x + 1 )} - \frac{1}{3 ( x + 1 ) ^{2}} + \frac{- x + 4}{9 ( x ^{2} + 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

= 2 (\int \frac{1}{9 ( x + 1 )} d x - \int \frac{1}{3 ( x + 1 ) ^{2}} d x + \int \frac{- x + 4}{9 ( x ^{2} + 2 )} d x)

\int \frac{1}{9 ( x + 1 )} d x = \frac{1}{9} ln ∣ x + 1 ∣

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

\int \frac{- x + 4}{9 ( x ^{2} + 2 )} d x = \frac{1}{9} (- \frac{1}{2} ln x^{2} + 2 + 22 arctan (\frac{x}{2}))

= 2 (\frac{1}{9} ln ∣ x + 1 ∣ - (- \frac{1}{3 ( x + 1 )}) + \frac{1}{9} (- \frac{1}{2} ln x^{2} + 2 + 22 arctan (\frac{x}{2})))

Vereinfache

= 2 (\frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{3 ( x + 1 )} + \frac{1}{9} (- \frac{1}{2} ln x^{2} + 2 + 22 arctan (\frac{x}{2})))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{9} ln ∣ x + 1 ∣ + \frac{1}{3 ( x + 1 )} + \frac{1}{9} (- \frac{1}{2} ln x^{2} + 2 + 22 arctan (\frac{x}{2}))) + C

in t e g r a l 2 (x + 1)

in t e g r a l - 2 x

in t e g r a l (e^{x - m})^{2}

in t e g r a l 2 - (x - 2)^{2}

in t e g r a l e^{- 2} t