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

Lösung

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

x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ - 4 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

Füge eine Konstante zur Lösung hinzu

= x - ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ - 4 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) + C

\int e^{\frac{- x}{5}} d x

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