integral of ((4x-5))/(x^2+8x+7)

Lösung

\int \frac{( 4 x - 5 )}{x ^{2} + 8 x + 7} d x

2 ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 21 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

Schritte zur Lösung

\int \frac{4 x - 5}{x ^{2} + 8 x + 7} d x

Vervollständige das Quadrat

x^{2} + 8 x + 7 : (x + 4)^{2} - 9

= \int \frac{4 x - 5}{( x + 4 ) ^{2} - 9} d x

Wende U-Substitution an

= \int \frac{4 u - 21}{u ^{2} - 9} d u

Faktorisiere

\frac{4 u - 21}{u ^{2} - 9}

= \int \frac{4 u - 21}{( u + 3 ) ( u - 3 )} d u

Multipliziere aus

\frac{4 u - 21}{( u + 3 ) ( u - 3 )} : \frac{4 u}{( u + 3 ) ( u - 3 )} - \frac{21}{( u + 3 ) ( u - 3 )}

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{4 u}{( u + 3 ) ( u - 3 )} d u - \int \frac{21}{( u + 3 ) ( u - 3 )} d u

\int \frac{4 u}{( u + 3 ) ( u - 3 )} d u = 2 ln (∣ u + 3 ∣ ∣ u - 3 ∣)

\int \frac{21}{( u + 3 ) ( u - 3 )} d u = 21 (- \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣)

= 2 ln (∣ u + 3 ∣ ∣ u - 3 ∣) - 21 (- \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣)

Setze in

u = x + 4

ein

= 2 ln (∣ x + 4 + 3 ∣ ∣ x + 4 - 3 ∣) - 21 (- \frac{1}{6} ln ∣ x + 4 + 3 ∣ + \frac{1}{6} ln ∣ x + 4 - 3 ∣)

Vereinfache

2 ln (∣ x + 4 + 3 ∣ ∣ x + 4 - 3 ∣) - 21 (- \frac{1}{6} ln ∣ x + 4 + 3 ∣ + \frac{1}{6} ln ∣ x + 4 - 3 ∣) : 2 ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 21 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

= 2 ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 21 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= 2 ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 21 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

\int 8 x + 3 sin (x) d x

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