integral of ((x-4))/(sqrt(x^2-4x))

Lösung

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

x^{2} - 4 x - 2 ln (\frac{1}{2} x^{2} - 4 x + x - 2) + C

Schritte zur Lösung

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

Multipliziere aus

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

= \int \frac{x}{x ^{2} - 4 x} - \frac{4}{x ^{2} - 4 x} d 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 ^{2} - 4 x} d x - \int \frac{4}{x ^{2} - 4 x} d x

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

= x^{2} - 4 x - 2 ln (\frac{1}{2} x^{2} - 4 x + x - 2) + C

\int \frac{x ^{3}}{3} cos (x^{4}) d x

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

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

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

\int (\frac{u ^{3} + 2 u ^{2} - u}{3 u}) d u