integral of 1/((x^2-9)^2)

Lösung

\int \frac{1}{( x ^{2} - 9 ) ^{2}} d x

\frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )} + C

Schritte zur Lösung

\int \frac{1}{( x ^{2} - 9 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{1}{( x ^{2} - 9 ) ^{2}} : \frac{1}{108 ( x + 3 )} + \frac{1}{36 ( x + 3 ) ^{2}} - \frac{1}{108 ( x - 3 )} + \frac{1}{36 ( x - 3 ) ^{2}}

= \int \frac{1}{108 ( x + 3 )} + \frac{1}{36 ( x + 3 ) ^{2}} - \frac{1}{108 ( x - 3 )} + \frac{1}{36 ( x - 3 ) ^{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}{108 ( x + 3 )} d x + \int \frac{1}{36 ( x + 3 ) ^{2}} d x - \int \frac{1}{108 ( x - 3 )} d x + \int \frac{1}{36 ( x - 3 ) ^{2}} d x

\int \frac{1}{108 ( x + 3 )} d x = \frac{1}{108} ln ∣ x + 3 ∣

\int \frac{1}{36 ( x + 3 ) ^{2}} d x = - \frac{1}{36 ( x + 3 )}

\int \frac{1}{108 ( x - 3 )} d x = \frac{1}{108} ln ∣ x - 3 ∣

\int \frac{1}{36 ( x - 3 ) ^{2}} d x = - \frac{1}{36 ( x - 3 )}

= \frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{108} ln ∣ x + 3 ∣ - \frac{1}{36 ( x + 3 )} - \frac{1}{108} ln ∣ x - 3 ∣ - \frac{1}{36 ( x - 3 )} + C

\frac{d}{d x} (3 x - 8)

t an g e n t f (x) = (1 + 3 x)^{10}, at x = 0

x \to 2 lim (x^{2} + 4 x)

\frac{d}{d x} ((2 x - 2)^{3} (3 x - 1)^{2})

x \to 0 + lim (\frac{1 - ln ( x )}{e ^{\frac{1}{x}}})