integral of (5x^2-x+1)/(x^3-4x)

Lösung

\int \frac{5 x ^{2} - x + 1}{x ^{3} - 4 x} d x

\frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + C

Schritte zur Lösung

\int \frac{5 x ^{2} - x + 1}{x ^{3} - 4 x} d x

Multipliziere aus

\frac{5 x ^{2} - x + 1}{x ^{3} - 4 x} : \frac{5 x ^{2}}{x ^{3} - 4 x} - \frac{x}{x ^{3} - 4 x} + \frac{1}{x ^{3} - 4 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{5 x ^{2}}{x ^{3} - 4 x} d x - \int \frac{x}{x ^{3} - 4 x} d x + \int \frac{1}{x ^{3} - 4 x} d x

\int \frac{5 x ^{2}}{x ^{3} - 4 x} d x = \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣)

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

\int \frac{1}{x ^{3} - 4 x} d x = - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣

= \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - (- \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣

Vereinfache

= \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + C

\int \frac{4}{x ln ( 3 x )} d x

\int \frac{1}{7 x} d x

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

\int cos (θ) tan (θ) d θ

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