integral of y/(y+1)

Lösung

\int \frac{y}{y + 1} d y

y + 1 - ln ∣ y + 1 ∣ + C

Schritte zur Lösung

\int \frac{y}{y + 1} d y

Wende U-Substitution an

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

Multipliziere aus

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u - ln ∣ u ∣

Setze in

u = y + 1

ein

= y + 1 - ln ∣ y + 1 ∣

Füge eine Konstante zur Lösung hinzu

= y + 1 - ln ∣ y + 1 ∣ + C

x y (\frac{d y}{d x}) = 3

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\int_{0}^{\frac{π}{4}} (tan^{2} (x)) d x