integral of (5x-2)/((x-1)^2(2x+1))

Lösung

\int \frac{5 x - 2}{( x - 1 ) ^{2} ( 2 x + 1 )} d x

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

Schritte zur Lösung

\int \frac{5 x - 2}{( x - 1 ) ^{2} ( 2 x + 1 )} d x

Ermittle den Partialbruch von

\frac{5 x - 2}{( x - 1 ) ^{2} ( 2 x + 1 )} : \frac{1}{x - 1} + \frac{1}{( x - 1 ) ^{2}} - \frac{2}{2 x + 1}

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

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

\int \frac{2}{2 x + 1} d x = ln ∣ 2 x + 1 ∣

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

Füge eine Konstante zur Lösung hinzu

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

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