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

Lösung

integral

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

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

Schritte zur Lösung

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

Multipliziere aus

\frac{x ^{3} - 3 x}{1 - x ^{4}} : \frac{x ^{3}}{1 - x ^{4}} - \frac{3 x}{1 - x ^{4}}

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

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

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

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

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} ln 1 - x^{4} - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1) + C

in t e g r a l \frac{x ^{3}}{5}

in t e g r a l cos^{2} (x) - 1

in t e g r a l \frac{x}{x + 3}

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

in t e g r a l y = ∣ x - 1 ∣