integral of (5x-11)/(x^2+10x+9)

Lösung

\int \frac{5 x - 11}{x ^{2} + 10 x + 9} d x

\frac{5}{2} ln (∣ x + 9 ∣ ∣ x + 1 ∣) - 36 (- \frac{1}{8} ln ∣ x + 9 ∣ + \frac{1}{8} ln ∣ x + 1 ∣) + C

Schritte zur Lösung

\int \frac{5 x - 11}{x ^{2} + 10 x + 9} d x

Vervollständige das Quadrat

x^{2} + 10 x + 9 : (x + 5)^{2} - 16

= \int \frac{5 x - 11}{( x + 5 ) ^{2} - 16} d x

Wende U-Substitution an

= \int \frac{5 u - 36}{u ^{2} - 16} d u

Faktorisiere

\frac{5 u - 36}{u ^{2} - 16}

= \int \frac{5 u - 36}{( u + 4 ) ( u - 4 )} d u

Multipliziere aus

\frac{5 u - 36}{( u + 4 ) ( u - 4 )} : \frac{5 u}{( u + 4 ) ( u - 4 )} - \frac{36}{( u + 4 ) ( u - 4 )}

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{5 u}{( u + 4 ) ( u - 4 )} d u - \int \frac{36}{( u + 4 ) ( u - 4 )} d u

\int \frac{5 u}{( u + 4 ) ( u - 4 )} d u = \frac{5}{2} ln (∣ u + 4 ∣ ∣ u - 4 ∣)

\int \frac{36}{( u + 4 ) ( u - 4 )} d u = 36 (- \frac{1}{8} ln ∣ u + 4 ∣ + \frac{1}{8} ln ∣ u - 4 ∣)

= \frac{5}{2} ln (∣ u + 4 ∣ ∣ u - 4 ∣) - 36 (- \frac{1}{8} ln ∣ u + 4 ∣ + \frac{1}{8} ln ∣ u - 4 ∣)

Setze in

u = x + 5

ein

= \frac{5}{2} ln (∣ x + 5 + 4 ∣ ∣ x + 5 - 4 ∣) - 36 (- \frac{1}{8} ln ∣ x + 5 + 4 ∣ + \frac{1}{8} ln ∣ x + 5 - 4 ∣)

Vereinfache

\frac{5}{2} ln (∣ x + 5 + 4 ∣ ∣ x + 5 - 4 ∣) - 36 (- \frac{1}{8} ln ∣ x + 5 + 4 ∣ + \frac{1}{8} ln ∣ x + 5 - 4 ∣) : \frac{5}{2} ln (∣ x + 9 ∣ ∣ x + 1 ∣) - 36 (- \frac{1}{8} ln ∣ x + 9 ∣ + \frac{1}{8} ln ∣ x + 1 ∣)

= \frac{5}{2} ln (∣ x + 9 ∣ ∣ x + 1 ∣) - 36 (- \frac{1}{8} ln ∣ x + 9 ∣ + \frac{1}{8} ln ∣ x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{5}{2} ln (∣ x + 9 ∣ ∣ x + 1 ∣) - 36 (- \frac{1}{8} ln ∣ x + 9 ∣ + \frac{1}{8} ln ∣ x + 1 ∣) + C

\int \frac{1}{( x ^{2} + 49 ) ^{3}} d x

\int \frac{1}{u ( u + 1 )} d u

\int \frac{1}{3 x ^{2} - 5} d x

\int \frac{8 x - 5}{3 x ^{2} - 5 x + 2} d x

\int \frac{3 a}{4 t} d t