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Beliebt Rechnen >

integral of (15)/(x(x^2+4)^2)

Lösung

\int \frac{15}{x ( x ^{2} + 4 ) ^{2}} d x

Lösung

\frac{15}{2} (\frac{1}{8} ln (x) - \frac{1}{16} ln x^{2} + 4 + \frac{1}{4 ( x ^{2} + 4 )}) + C

Schritte zur Lösung

\int \frac{15}{x ( x ^{2} + 4 ) ^{2}} d x

Entferne die Konstante:

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

= 15 \cdot \int \frac{1}{x ( x ^{2} + 4 ) ^{2}} d x

Wende U-Substitution an

= 15 \cdot \int \frac{1}{2 ( u - 4 ) u ^{2}} d u

Entferne die Konstante:

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

= 15 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 4 ) u ^{2}} d u

Ermittle den Partialbruch von

\frac{1}{( u - 4 ) u ^{2}} : \frac{1}{16 ( u - 4 )} - \frac{1}{16 u} - \frac{1}{4 u ^{2}}

= 15 \cdot \frac{1}{2} \cdot \int \frac{1}{16 ( u - 4 )} - \frac{1}{16 u} - \frac{1}{4 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

= 15 \cdot \frac{1}{2} (\int \frac{1}{16 ( u - 4 )} d u - \int \frac{1}{16 u} d u - \int \frac{1}{4 u ^{2}} d u)

\int \frac{1}{16 ( u - 4 )} d u = \frac{1}{16} ln ∣ u - 4 ∣

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

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

= 15 \cdot \frac{1}{2} (\frac{1}{16} ln ∣ u - 4 ∣ - \frac{1}{16} ln ∣ u ∣ - (- \frac{1}{4 u}))

Setze in

u = x^{2} + 4

ein

= 15 \cdot \frac{1}{2} (\frac{1}{16} ln x^{2} + 4 - 4 - \frac{1}{16} ln x^{2} + 4 - (- \frac{1}{4 ( x ^{2} + 4 )}))

Vereinfache

15 \cdot \frac{1}{2} (\frac{1}{16} ln x^{2} + 4 - 4 - \frac{1}{16} ln x^{2} + 4 - (- \frac{1}{4 ( x ^{2} + 4 )})) : \frac{15}{2} (\frac{1}{8} ln (x) - \frac{1}{16} ln x^{2} + 4 + \frac{1}{4 ( x ^{2} + 4 )})

= \frac{15}{2} (\frac{1}{8} ln (x) - \frac{1}{16} ln x^{2} + 4 + \frac{1}{4 ( x ^{2} + 4 )})

Füge eine Konstante zur Lösung hinzu

= \frac{15}{2} (\frac{1}{8} ln (x) - \frac{1}{16} ln x^{2} + 4 + \frac{1}{4 ( x ^{2} + 4 )}) + C

Graph

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