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

Lösung

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

ln x^{2} - x + x + C

Schritte zur Lösung

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

\frac{x ^{2} + x - 1}{x ^{2} - x} = \frac{2 x - 1}{x ^{2} - x} + 1

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

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

\int 1 d x = x

= ln x^{2} - x + x

Füge eine Konstante zur Lösung hinzu

= ln x^{2} - x + x + C

\frac{d}{d x} (\frac{3 x ^{2}}{2} - \frac{7}{5 x ^{2}})

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

t an g e n t y = 3 x^{2} - 4 x + 5

d er i v a t i v e y = ln (x) x

x \to 5 lim (\frac{5}{( x - 5 ) ^{2}})