integral of 1/((x^2+6x+5)^2)

Lösung

\int \frac{1}{( x ^{2} + 6 x + 5 ) ^{2}} d x

\frac{1}{32} ln ∣ x + 5 ∣ - \frac{1}{16 ( x + 5 )} - \frac{1}{32} ln ∣ x + 1 ∣ - \frac{1}{16 ( x + 1 )} + C

Schritte zur Lösung

\int \frac{1}{( x ^{2} + 6 x + 5 ) ^{2}} d x

Vervollständige das Quadrat

x^{2} + 6 x + 5 : (x + 3)^{2} - 4

= \int \frac{1}{( ( x + 3 ) ^{2} - 4 ) ^{2}} d x

Wende U-Substitution an

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

Ermittle den Partialbruch von

\frac{1}{( u ^{2} - 4 ) ^{2}} : \frac{1}{32 ( u + 2 )} + \frac{1}{16 ( u + 2 ) ^{2}} - \frac{1}{32 ( u - 2 )} + \frac{1}{16 ( u - 2 ) ^{2}}

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

\int \frac{1}{32 ( u + 2 )} d u = \frac{1}{32} ln ∣ u + 2 ∣

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

\int \frac{1}{32 ( u - 2 )} d u = \frac{1}{32} ln ∣ u - 2 ∣

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

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

Setze in

u = x + 3

ein

= \frac{1}{32} ln ∣ x + 3 + 2 ∣ - \frac{1}{16 ( x + 3 + 2 )} - \frac{1}{32} ln ∣ x + 3 - 2 ∣ - \frac{1}{16 ( x + 3 - 2 )}

Vereinfache

\frac{1}{32} ln ∣ x + 3 + 2 ∣ - \frac{1}{16 ( x + 3 + 2 )} - \frac{1}{32} ln ∣ x + 3 - 2 ∣ - \frac{1}{16 ( x + 3 - 2 )} : \frac{1}{32} ln ∣ x + 5 ∣ - \frac{1}{16 ( x + 5 )} - \frac{1}{32} ln ∣ x + 1 ∣ - \frac{1}{16 ( x + 1 )}

= \frac{1}{32} ln ∣ x + 5 ∣ - \frac{1}{16 ( x + 5 )} - \frac{1}{32} ln ∣ x + 1 ∣ - \frac{1}{16 ( x + 1 )}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{32} ln ∣ x + 5 ∣ - \frac{1}{16 ( x + 5 )} - \frac{1}{32} ln ∣ x + 1 ∣ - \frac{1}{16 ( x + 1 )} + C

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