integral of (x^4)/((1-x^2)^3)

Lösung

\int \frac{x ^{4}}{( 1 - x ^{2} ) ^{3}} d x

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

Schritte zur Lösung

\int \frac{x ^{4}}{( 1 - x ^{2} ) ^{3}} d x

Ermittle den Partialbruch von

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

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

\int \frac{3}{16 ( x + 1 )} d x = \frac{3}{16} ln ∣ x + 1 ∣

\int \frac{5}{16 ( x + 1 ) ^{2}} d x = - \frac{5}{16 ( x + 1 )}

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

\int \frac{3}{16 ( x - 1 )} d x = \frac{3}{16} ln ∣ x - 1 ∣

\int \frac{5}{16 ( x - 1 ) ^{2}} d x = - \frac{5}{16 ( x - 1 )}

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{x ^{3} ln ( x )}{2} d x

\int (x^{5} - x^{3} + x) e^{- x} d x

\int x ln (3 x) - x d x

\int \frac{1}{24} d x

\int \frac{1}{4 sec ( x )} d x