integral of (3x-4)/(x^2-3x+2)

Lösung

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

\frac{3}{2} ln 4 x^{2} - 12 x + 8 - \frac{1}{2} ln ∣ 2 x - 2 ∣ + \frac{1}{2} ln ∣ 2 x - 4 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

\frac{3 x - 4}{x ^{2} - 3 x + 2} : \frac{3 x}{x ^{2} - 3 x + 2} - \frac{4}{x ^{2} - 3 x + 2}

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

\int \frac{3 x}{x ^{2} - 3 x + 2} d x = 6 (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣))

\int \frac{4}{x ^{2} - 3 x + 2} d x = - 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)

= 6 (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣)) - (- 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣))

Vereinfache

6 (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣)) - (- 8 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)) : \frac{3}{2} ln 4 x^{2} - 12 x + 8 - \frac{1}{2} ln ∣ 2 x - 2 ∣ + \frac{1}{2} ln ∣ 2 x - 4 ∣

= \frac{3}{2} ln 4 x^{2} - 12 x + 8 - \frac{1}{2} ln ∣ 2 x - 2 ∣ + \frac{1}{2} ln ∣ 2 x - 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} ln 4 x^{2} - 12 x + 8 - \frac{1}{2} ln ∣ 2 x - 2 ∣ + \frac{1}{2} ln ∣ 2 x - 4 ∣ + C

a re a x + y^{2} = 42, x + y = 0

a re a x = 25 - y^{2}, x = y^{2} - 25

\frac{d}{d x} (1 0^{- 6} x^{3})

d er i v a t i v e f^{^{'}} (3) - \frac{3}{8 ( x + 4 ) ^{4}}

\frac{d}{d x} (x ln (x + 1))