integral of-1/(Ux)+1/(U(x-U))

Lösung

\int - \frac{1}{Ux} + \frac{1}{U ( x - U )} d x

\frac{1}{U} ln ∣ - U + x ∣ - \frac{1}{U} ln ∣ x ∣ + C

Schritte zur Lösung

\int - \frac{1}{Ux} + \frac{1}{U ( x - U )} 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}{( - U + x ) U} d x - \int \frac{1}{Ux} d x

\int \frac{1}{( - U + x ) U} d x = \frac{1}{U} ln ∣ - U + x ∣

\int \frac{1}{Ux} d x = \frac{1}{U} ln ∣ x ∣

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

Füge eine Konstante zur Lösung hinzu

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

\frac{\partial}{\partial y} (\frac{e ^{y}}{x + y ^{2}})

t an g e n t \frac{6}{3 x ^{2}} - x, at x = 1

\frac{d y}{d x} = 17

\frac{\partial}{\partial x} (y cos (π x))

\int t^{6} ln (t) d t