integral v/((v-1)^2)

Lösung

integral

\frac{v}{( v - 1 ) ^{2}}

ln ∣ v - 1 ∣ - \frac{1}{v - 1} + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{u + 1}{u ^{2}} d u

Multipliziere aus

\frac{u + 1}{u ^{2}} : \frac{1}{u} + \frac{1}{u ^{2}}

= \int \frac{1}{u} + \frac{1}{u ^{2}} d u

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} d u + \int \frac{1}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

= ln ∣ u ∣ - \frac{1}{u}

Setze in

u = v - 1

ein

= ln ∣ v - 1 ∣ - \frac{1}{v - 1}

Füge eine Konstante zur Lösung hinzu

= ln ∣ v - 1 ∣ - \frac{1}{v - 1} + C

in t e g r a l (4 + 16 e^{8} x)^{\frac{1}{2}}

in t e g r a l \frac{( 1 - x )}{x ^{2}}

in t e g r a l \frac{e ^{- 1}}{x}

in t e g r a l (x - 1)^{2}

in t e g r a l (3^{x} + 3^{- x})^{2}