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Beliebt Rechnen >

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

Lösung

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

Lösung

\frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )} + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

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

= \frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )} + C

Graph

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