integral of (x^3)/((x^2-4)^2)

Lösung

\int \frac{x ^{3}}{( x ^{2} - 4 ) ^{2}} d x

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

Schritte zur Lösung

\int \frac{x ^{3}}{( x ^{2} - 4 ) ^{2}} d x

Wende integrale Substitution an

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

Ermittle den Partialbruch von

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

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

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

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

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

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

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

= \frac{1}{2} ln ∣ u + 1 ∣ - (- \frac{1}{4 ( u + 1 )}) + \frac{1}{2} ln ∣ u - 1 ∣ - \frac{1}{4 ( u - 1 )}

Setze in

u = \frac{x}{2}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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