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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

Addiere gleiche Elemente:

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

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

Füge eine Konstante zur Lösung hinzu

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

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

x \to 0 lim (2 x + x)

\int e^{- 2 θ} cos (θ) d θ

x \to 0 lim (x cos (\frac{2}{x}))

x \to 1 + lim (\frac{8}{1 - x})