integral of (2x^2+3x+7)/(x^3+x^2-x-1)

Lösung

\int \frac{2 x ^{2} + 3 x + 7}{x ^{3} + x ^{2} - x - 1} d x

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

Schritte zur Lösung

\int \frac{2 x ^{2} + 3 x + 7}{x ^{3} + x ^{2} - x - 1} d x

Multipliziere aus

\frac{2 x ^{2} + 3 x + 7}{x ^{3} + x ^{2} - x - 1} : \frac{2 x ^{2}}{x ^{3} + x ^{2} - x - 1} + \frac{3 x}{x ^{3} + x ^{2} - x - 1} + \frac{7}{x ^{3} + x ^{2} - x - 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 ^{2}}{x ^{3} + x ^{2} - x - 1} d x + \int \frac{3 x}{x ^{3} + x ^{2} - x - 1} d x + \int \frac{7}{x ^{3} + x ^{2} - x - 1} d x

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

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

\int \frac{7}{x ^{3} + x ^{2} - x - 1} d x = 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

= \frac{3}{2} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{2} ln ∣ x - 1 ∣ + 3 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

Vereinfache

\frac{3}{2} ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{2} ln ∣ x - 1 ∣ + 3 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) : - ln ∣ x + 1 ∣ + 3 ln ∣ x - 1 ∣ + \frac{3}{x + 1}

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

Füge eine Konstante zur Lösung hinzu

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

\int x \cdot arcsec (x) d x

\int \frac{sec ^{6} ( 2 r )}{tan ^{2} ( 2 r )} d r

\int 2 x sin^{2} (x^{2}) d x

\int \frac{x}{( 4 x + 9 ) ^{2}} d x

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