integral ((x^4-4x^2+x+1))/((x^2-4))

Lösung

integral

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

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

Schritte zur Lösung

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

Lange Division

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

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

Füge eine Konstante zur Lösung hinzu

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

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

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

in t e g r a l sin^{3} (2 x + 5)

in t e g r a l x^{2} lo g_{10} (1 + x)

in t e g r a l x^{2} e^{l n (x) + x}