integral of (2+2x)/(x-4x(1+x))

Lösung

\int \frac{2 + 2 x}{x - 4 x ( 1 + x )} d x

\frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣) + C

Schritte zur Lösung

\int \frac{2 + 2 x}{x - 4 x ( 1 + x )} d x

Multipliziere aus

\frac{2 + 2 x}{x - 4 x ( 1 + x )} : \frac{2}{x - 4 x ( 1 + x )} + \frac{2 x}{x - 4 x ( 1 + 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{2}{x - 4 x ( 1 + x )} d x + \int \frac{2 x}{x - 4 x ( 1 + x )} d x

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

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

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

Vereinfache

2 (\frac{1}{3} ln ∣ 4 x + 3 ∣ - \frac{1}{3} ln (4 ∣ x ∣)) - \frac{1}{2} ln ∣ 4 x + 3 ∣ : \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣)

= \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣) + C

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

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

\int_{1}^{2} (2 x - 4)^{4} d x

y^{^{'}} - \frac{2}{t} y = t^{2} e^{t}, y (1) = 3

d er i v a t i v e 5^{x^{2}}