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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

= \frac{1}{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}}

= \frac{1}{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

= \frac{1}{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}

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

Setze in

u = 1 + x^{2}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{x ^{3}}{x ^{2} - 25} d x

x \to 0 lim (\frac{x ^{2} - 6}{x})

n or ma l f (x) = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

\int \frac{2 x}{x ^{2} - 2 x + 5} d x

x \to 0 lim (\frac{( x + 64 ) ^{\frac{1}{3}} - 4}{x})