integral of (x^3+x^2)/(x^2-9x+20)

Lösung

\int \frac{x ^{3} + x ^{2}}{x ^{2} - 9 x + 20} d x

\frac{x ^{2}}{2} + 10 x + \frac{9}{2} ln 4 x^{2} - 36 x + 80 - \frac{41}{2} ln ∣ 2 x - 8 ∣ + \frac{41}{2} ln ∣ 2 x - 10 ∣ - 64 ln ∣ x - 4 ∣ + 125 ln ∣ x - 5 ∣ + C

Schritte zur Lösung

\int \frac{x ^{3} + x ^{2}}{x ^{2} - 9 x + 20} d x

Multipliziere aus

\frac{x ^{3} + x ^{2}}{x ^{2} - 9 x + 20} : \frac{x ^{3}}{x ^{2} - 9 x + 20} + \frac{x ^{2}}{x ^{2} - 9 x + 20}

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{x ^{3}}{x ^{2} - 9 x + 20} d x + \int \frac{x ^{2}}{x ^{2} - 9 x + 20} d x

\int \frac{x ^{3}}{x ^{2} - 9 x + 20} d x = \frac{x ^{2}}{2} + 9 x - 64 ln ∣ x - 4 ∣ + 125 ln ∣ x - 5 ∣

\int \frac{x ^{2}}{x ^{2} - 9 x + 20} d x = 4 (\frac{9}{8} ln 4 x^{2} - 36 x + 80 - \frac{41}{8} ln ∣ 2 x - 8 ∣ + \frac{41}{8} ln ∣ 2 x - 10 ∣) + x

= \frac{x ^{2}}{2} + 9 x - 64 ln ∣ x - 4 ∣ + 125 ln ∣ x - 5 ∣ + 4 (\frac{9}{8} ln 4 x^{2} - 36 x + 80 - \frac{41}{8} ln ∣ 2 x - 8 ∣ + \frac{41}{8} ln ∣ 2 x - 10 ∣) + x

Vereinfache

\frac{x ^{2}}{2} + 9 x - 64 ln ∣ x - 4 ∣ + 125 ln ∣ x - 5 ∣ + 4 (\frac{9}{8} ln 4 x^{2} - 36 x + 80 - \frac{41}{8} ln ∣ 2 x - 8 ∣ + \frac{41}{8} ln ∣ 2 x - 10 ∣) + x : \frac{x ^{2}}{2} + 10 x + \frac{9}{2} ln 4 x^{2} - 36 x + 80 - \frac{41}{2} ln ∣ 2 x - 8 ∣ + \frac{41}{2} ln ∣ 2 x - 10 ∣ - 64 ln ∣ x - 4 ∣ + 125 ln ∣ x - 5 ∣

= \frac{x ^{2}}{2} + 10 x + \frac{9}{2} ln 4 x^{2} - 36 x + 80 - \frac{41}{2} ln ∣ 2 x - 8 ∣ + \frac{41}{2} ln ∣ 2 x - 10 ∣ - 64 ln ∣ x - 4 ∣ + 125 ln ∣ x - 5 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{2}}{2} + 10 x + \frac{9}{2} ln 4 x^{2} - 36 x + 80 - \frac{41}{2} ln ∣ 2 x - 8 ∣ + \frac{41}{2} ln ∣ 2 x - 10 ∣ - 64 ln ∣ x - 4 ∣ + 125 ln ∣ x - 5 ∣ + C

a re a 6 - x^{2}, x^{2} - 2, 0, 3

\int (\frac{π}{2} - 5 x^{- 0.5}) d x

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

\int \frac{e ^{- x}}{e ^{x} - 1} d x

y^{^{'}} = - \frac{4 t}{y}