integral of (32x^2)/((x-6)(x+2)^2)

Lösung

\int \frac{32 x ^{2}}{( x - 6 ) ( x + 2 ) ^{2}} d x

32 (\frac{7}{16} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )} + \frac{9}{16} ln ∣ x - 6 ∣) + C

Schritte zur Lösung

\int \frac{32 x ^{2}}{( x - 6 ) ( x + 2 ) ^{2}} d x

Entferne die Konstante:

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

= 32 \cdot \int \frac{x ^{2}}{( x - 6 ) ( x + 2 ) ^{2}} d x

Wende U-Substitution an

= 32 \cdot \int \frac{( u - 2 ) ^{2}}{u ^{2} ( u - 8 )} d u

Ermittle den Partialbruch von

\frac{( u - 2 ) ^{2}}{u ^{2} ( u - 8 )} : \frac{7}{16 u} - \frac{1}{2 u ^{2}} + \frac{9}{16 ( u - 8 )}

= 32 \cdot \int \frac{7}{16 u} - \frac{1}{2 u ^{2}} + \frac{9}{16 ( u - 8 )} 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

= 32 (\int \frac{7}{16 u} d u - \int \frac{1}{2 u ^{2}} d u + \int \frac{9}{16 ( u - 8 )} d u)

\int \frac{7}{16 u} d u = \frac{7}{16} ln ∣ u ∣

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

\int \frac{9}{16 ( u - 8 )} d u = \frac{9}{16} ln ∣ u - 8 ∣

= 32 (\frac{7}{16} ln ∣ u ∣ - (- \frac{1}{2 u}) + \frac{9}{16} ln ∣ u - 8 ∣)

Setze in

u = x + 2

ein

= 32 (\frac{7}{16} ln ∣ x + 2 ∣ - (- \frac{1}{2 ( x + 2 )}) + \frac{9}{16} ln ∣ x + 2 - 8 ∣)

Vereinfache

32 (\frac{7}{16} ln ∣ x + 2 ∣ - (- \frac{1}{2 ( x + 2 )}) + \frac{9}{16} ln ∣ x + 2 - 8 ∣) : 32 (\frac{7}{16} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )} + \frac{9}{16} ln ∣ x - 6 ∣)

= 32 (\frac{7}{16} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )} + \frac{9}{16} ln ∣ x - 6 ∣)

Füge eine Konstante zur Lösung hinzu

= 32 (\frac{7}{16} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )} + \frac{9}{16} ln ∣ x - 6 ∣) + C

\int x^{3} x^{2} + 26 d x

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