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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int (7 x^{2} + 5 x^{- 2}) d x

\int x^{2} (5 x^{3} - 2)^{4} d x

\int \frac{1}{18 - x ^{2}} d x

\int csc (z) d z

\int 5^{x + 3} d x