integral of (4x^2)/(x^3+x^2-x-1)

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Ermittle den Partialbruch von

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

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

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

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e f (x) = \frac{1}{5 - 2 e ^{x}}

\int \frac{3 x ^{2} + 4}{0.3 x ^{3} + 1.2 x} d x

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

\frac{d}{d x} (ln (cot (x)))

x \to - 5 - lim (x^{2} + \frac{4}{x + 5})