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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

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

Füge eine Konstante zur Lösung hinzu

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