integral of (2x^3-6x^2-10x-6)/(x^2-1)

Lösung

\int \frac{2 x ^{3} - 6 x ^{2} - 10 x - 6}{x ^{2} - 1} d x

x^{2} - 6 x - 4 ln x^{2} - 1 + 6 ln ∣ x + 1 ∣ - 6 ln ∣ x - 1 ∣ - 1 + C

Schritte zur Lösung

\int \frac{2 x ^{3} - 6 x ^{2} - 10 x - 6}{x ^{2} - 1} d x

Multipliziere aus

\frac{2 x ^{3} - 6 x ^{2} - 10 x - 6}{x ^{2} - 1} : \frac{2 x ^{3}}{x ^{2} - 1} - \frac{6 x ^{2}}{x ^{2} - 1} - \frac{10 x}{x ^{2} - 1} - \frac{6}{x ^{2} - 1}

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{2 x ^{3}}{x ^{2} - 1} d x - \int \frac{6 x ^{2}}{x ^{2} - 1} d x - \int \frac{10 x}{x ^{2} - 1} d x - \int \frac{6}{x ^{2} - 1} d x

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

\int \frac{6 x ^{2}}{x ^{2} - 1} d x = - 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)

\int \frac{10 x}{x ^{2} - 1} d x = 5 ln x^{2} - 1

\int \frac{6}{x ^{2} - 1} d x = - 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= x^{2} - 1 + ln x^{2} - 1 - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)) - 5 ln x^{2} - 1 - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣))

Vereinfache

x^{2} - 1 + ln x^{2} - 1 - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)) - 5 ln x^{2} - 1 - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)) : x^{2} - 6 x - 4 ln x^{2} - 1 + 6 ln ∣ x + 1 ∣ - 6 ln ∣ x - 1 ∣ - 1

= x^{2} - 6 x - 4 ln x^{2} - 1 + 6 ln ∣ x + 1 ∣ - 6 ln ∣ x - 1 ∣ - 1

Füge eine Konstante zur Lösung hinzu

= x^{2} - 6 x - 4 ln x^{2} - 1 + 6 ln ∣ x + 1 ∣ - 6 ln ∣ x - 1 ∣ - 1 + C

3 y^{^{''}} - y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 0

\int \frac{tan ^{3} ( \frac{4}{z} )}{z ^{2}} d z

\int \frac{1}{e ^{\frac{x}{2}}} d x

\int \frac{sin ^{5} ( 2 x )}{cos ^{2} ( 2 x )} d x

\frac{d}{d y} (e^{2 x y})