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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

d er i v a t i v e - 4 sin (t)

n = 1 \sum \infty (\frac{1}{2})^{2 n}

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

\int \frac{e ^{x}}{1 + 16 e ^{2 x}} d x

\frac{d}{d x} (ln (3 \frac{x - 1}{x + 1}))