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

Lösung

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

\frac{1}{x} + ln ∣ x ∣ + C

Schritte zur Lösung

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

Multipliziere aus

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= - (- \frac{1}{x}) + ln ∣ x ∣

Vereinfache

= \frac{1}{x} + ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{x} + ln ∣ x ∣ + C

\frac{\partial}{\partial x} (3 x^{2} y^{4} - 4)

\int \frac{1}{sin ( x ) + cos ( x )} d x

\int_{- 1}^{1} \frac{1}{2 x + 3} d x

\frac{d}{d x} (\frac{1}{3} x^{(\frac{1}{2})} - x^{(\frac{3}{2})})

t an g e n t f (x) = \frac{1}{x}, at x = \frac{1}{4}