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

Lösung

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

- \frac{3}{16} ln ∣ x ∣ + \frac{17}{128} ln ∣ x + 2 ∣ - \frac{5}{32 ( x + 2 )} + \frac{1}{16 ( x + 2 ) ^{2}} + \frac{7}{128} ln ∣ x - 2 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{2 x + 3}{x ( x ^{2} - 4 ) ( x + 2 ) ^{2}} : - \frac{3}{16 x} + \frac{17}{128 ( x + 2 )} + \frac{5}{32 ( x + 2 ) ^{2}} - \frac{1}{8 ( x + 2 ) ^{3}} + \frac{7}{128 ( x - 2 )}

= \int - \frac{3}{16 x} + \frac{17}{128 ( x + 2 )} + \frac{5}{32 ( x + 2 ) ^{2}} - \frac{1}{8 ( x + 2 ) ^{3}} + \frac{7}{128 ( x - 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{3}{16 x} d x + \int \frac{17}{128 ( x + 2 )} d x + \int \frac{5}{32 ( x + 2 ) ^{2}} d x - \int \frac{1}{8 ( x + 2 ) ^{3}} d x + \int \frac{7}{128 ( x - 2 )} d x

\int \frac{3}{16 x} d x = \frac{3}{16} ln ∣ x ∣

\int \frac{17}{128 ( x + 2 )} d x = \frac{17}{128} ln ∣ x + 2 ∣

\int \frac{5}{32 ( x + 2 ) ^{2}} d x = - \frac{5}{32 ( x + 2 )}

\int \frac{1}{8 ( x + 2 ) ^{3}} d x = - \frac{1}{16 ( x + 2 ) ^{2}}

\int \frac{7}{128 ( x - 2 )} d x = \frac{7}{128} ln ∣ x - 2 ∣

= - \frac{3}{16} ln ∣ x ∣ + \frac{17}{128} ln ∣ x + 2 ∣ - \frac{5}{32 ( x + 2 )} - (- \frac{1}{16 ( x + 2 ) ^{2}}) + \frac{7}{128} ln ∣ x - 2 ∣

Vereinfache

= - \frac{3}{16} ln ∣ x ∣ + \frac{17}{128} ln ∣ x + 2 ∣ - \frac{5}{32 ( x + 2 )} + \frac{1}{16 ( x + 2 ) ^{2}} + \frac{7}{128} ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{3}{16} ln ∣ x ∣ + \frac{17}{128} ln ∣ x + 2 ∣ - \frac{5}{32 ( x + 2 )} + \frac{1}{16 ( x + 2 ) ^{2}} + \frac{7}{128} ln ∣ x - 2 ∣ + C

\frac{d}{d x} (25 x)

y (1 + x^{2}) y^{^{'}} - x (1 + y^{2}) = 0, y (0) = 3

x \to - 8 lim (12 x + 11)

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

n = 1 \sum \infty z^{n - 2}