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

Lösung

integral

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

- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )} + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

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

= - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )} + C

in t e g r a l e^{z} (z + 1)

in t e g r a l (x - 11)^{7}

in t e g r a l \frac{1}{x ^{2} ( x + 1 ) ^{2}}

in t e g r a l \frac{( x ^{2} )}{3}

in t e g r a l \frac{x}{( 0.5 x - 1 ) ^{0.5}}