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

Lösung

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

x - \frac{3}{2} ln ∣ x + 2 ∣ - \frac{1}{x + 2} + \frac{3}{2} ln ∣ x - 2 ∣ - \frac{1}{x - 2} + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{x ^{4}}{( x ^{2} - 4 ) ^{2}} : 1 - \frac{3}{2 ( x + 2 )} + \frac{1}{( x + 2 ) ^{2}} + \frac{3}{2 ( x - 2 )} + \frac{1}{( x - 2 ) ^{2}}

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

\int 1 d x = x

\int \frac{3}{2 ( x + 2 )} d x = \frac{3}{2} ln ∣ x + 2 ∣

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

\int \frac{3}{2 ( x - 2 )} d x = \frac{3}{2} ln ∣ x - 2 ∣

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

= x - \frac{3}{2} ln ∣ x + 2 ∣ - \frac{1}{x + 2} + \frac{3}{2} ln ∣ x - 2 ∣ - \frac{1}{x - 2}

Füge eine Konstante zur Lösung hinzu

= x - \frac{3}{2} ln ∣ x + 2 ∣ - \frac{1}{x + 2} + \frac{3}{2} ln ∣ x - 2 ∣ - \frac{1}{x - 2} + C

\int \frac{4 x ^{3} - 1}{x ^{4} - x + 4} d x

\int x (x - 4)^{9} d x

\int \frac{6 + e ^{3 x}}{e ^{2 x}} d x

\int \frac{2 x ^{2}}{x ^{2} - 6 x + 10} d x

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