integral of (2x^2+3)/(x^4-2x^2+1)

Lösung

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

\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} - \frac{3}{4 ( x + 1 )} - \frac{3}{4 ( x - 1 )} + C

Schritte zur Lösung

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

Multipliziere aus

\frac{2 x ^{2} + 3}{x ^{4} - 2 x ^{2} + 1} : \frac{2 x ^{2}}{x ^{4} - 2 x ^{2} + 1} + \frac{3}{x ^{4} - 2 x ^{2} + 1}

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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