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

Lösung

\int \frac{2 x ^{2} + x}{( x - 1 ) ^{2} ( x + 1 ) ^{2}} 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 )}) - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x + 1 )} + C

Schritte zur Lösung

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

Multipliziere aus

\frac{2 x ^{2} + x}{( x - 1 ) ^{2} ( x + 1 ) ^{2}} : \frac{2 x ^{2}}{( x - 1 ) ^{2} ( x + 1 ) ^{2}} + \frac{x}{( x - 1 ) ^{2} ( x + 1 ) ^{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{2 x ^{2}}{( x - 1 ) ^{2} ( x + 1 ) ^{2}} d x + \int \frac{x}{( x - 1 ) ^{2} ( x + 1 ) ^{2}} d x

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

Füge eine Konstante zur Lösung hinzu

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

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