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

Lösung

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

arctan (v) - \frac{1}{2} ln 1 + v^{2} + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

\int \frac{v}{1 + v ^{2}} d v = \frac{1}{2} ln 1 + v^{2}

= arctan (v) - \frac{1}{2} ln 1 + v^{2}

Füge eine Konstante zur Lösung hinzu

= arctan (v) - \frac{1}{2} ln 1 + v^{2} + C

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