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

Lösung

integral

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

\frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )} + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

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

= \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)) + \frac{1}{2} (\frac{x}{2 ( x ^{2} + 1 )} + \frac{1}{2} arctan (x) + \frac{1}{2 ( x ^{2} + 1 )})

Vereinfache

\frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)) + \frac{1}{2} (\frac{x}{2 ( x ^{2} + 1 )} + \frac{1}{2} arctan (x) + \frac{1}{2 ( x ^{2} + 1 )}) : \frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )}

= \frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )}

Füge eine Konstante zur Lösung hinzu

= \frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )} + C

in t e g r a l \frac{1}{( 2 + x )}

in t e g r a l cos (t^{3})

in t e g r a l (x + 9)^{4}

in t e g r a l \frac{1 - 1}{x ^{4}}^{5}

in t e g r a l \frac{( e ^{2} \cdot y )}{( 4 - y )}