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Beliebt Rechnen >

integral of (x^3)/(x^2+6x+5)

Lösung

\int \frac{x ^{3}}{x ^{2} + 6 x + 5} d x

Lösung

\frac{1}{2} (x + 3)^{2} - 9 (x + 3) + \frac{31}{2} ln (∣ x + 5 ∣ ∣ x + 1 ∣) - 63 (- \frac{1}{4} ln ∣ x + 5 ∣ + \frac{1}{4} ln ∣ x + 1 ∣) + C

Schritte zur Lösung

\int \frac{x ^{3}}{x ^{2} + 6 x + 5} d x

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Lange Division

\frac{( u - 3 ) ^{3}}{u ^{2} - 4} : u - 9 + \frac{31 u - 63}{u ^{2} - 4}

= \int u - 9 + \frac{31 u - 63}{u ^{2} - 4} 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 u d u - \int 9 d u + \int \frac{31 u - 63}{u ^{2} - 4} d u

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

\int 9 d u = 9 u

\int \frac{31 u - 63}{u ^{2} - 4} d u = \frac{31}{2} ln (∣ u + 2 ∣ ∣ u - 2 ∣) - 63 (- \frac{1}{4} ln ∣ u + 2 ∣ + \frac{1}{4} ln ∣ u - 2 ∣)

= \frac{u ^{2}}{2} - 9 u + \frac{31}{2} ln (∣ u + 2 ∣ ∣ u - 2 ∣) - 63 (- \frac{1}{4} ln ∣ u + 2 ∣ + \frac{1}{4} ln ∣ u - 2 ∣)

Setze in

u = x + 3

ein

= \frac{( x + 3 ) ^{2}}{2} - 9 (x + 3) + \frac{31}{2} ln (∣ x + 3 + 2 ∣ ∣ x + 3 - 2 ∣) - 63 (- \frac{1}{4} ln ∣ x + 3 + 2 ∣ + \frac{1}{4} ln ∣ x + 3 - 2 ∣)

Vereinfache

\frac{( x + 3 ) ^{2}}{2} - 9 (x + 3) + \frac{31}{2} ln (∣ x + 3 + 2 ∣ ∣ x + 3 - 2 ∣) - 63 (- \frac{1}{4} ln ∣ x + 3 + 2 ∣ + \frac{1}{4} ln ∣ x + 3 - 2 ∣) : \frac{1}{2} (x + 3)^{2} - 9 (x + 3) + \frac{31}{2} ln (∣ x + 5 ∣ ∣ x + 1 ∣) - 63 (- \frac{1}{4} ln ∣ x + 5 ∣ + \frac{1}{4} ln ∣ x + 1 ∣)

= \frac{1}{2} (x + 3)^{2} - 9 (x + 3) + \frac{31}{2} ln (∣ x + 5 ∣ ∣ x + 1 ∣) - 63 (- \frac{1}{4} ln ∣ x + 5 ∣ + \frac{1}{4} ln ∣ x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (x + 3)^{2} - 9 (x + 3) + \frac{31}{2} ln (∣ x + 5 ∣ ∣ x + 1 ∣) - 63 (- \frac{1}{4} ln ∣ x + 5 ∣ + \frac{1}{4} ln ∣ x + 1 ∣) + C

Graph

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