integral 1/(v(v+1))

Lösung

integral

\frac{1}{v ( v + 1 )}

ln ∣ v ∣ - ln ∣ v + 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{v ( v + 1 )} d v

Ermittle den Partialbruch von

\frac{1}{v ( v + 1 )} : \frac{1}{v} - \frac{1}{v + 1}

= \int \frac{1}{v} - \frac{1}{v + 1} d v

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}{v} d v - \int \frac{1}{v + 1} d v

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v + 1} d v = ln ∣ v + 1 ∣

= ln ∣ v ∣ - ln ∣ v + 1 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ v ∣ - ln ∣ v + 1 ∣ + C

in t e g r a l 1 - cos (2 x)

in t e g r a l t + 2

in t e g r a l x^{3} - 2 x + 5

in t e g r a l \frac{( x ^{3} )}{( x ^{2} + 2 x + 1 )}

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