integral of x/(1+xy)

Lösung

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

\frac{1}{y ^{2}} (1 + y x - ln ∣ 1 + y x ∣) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Multipliziere aus

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

= \frac{1}{y ^{2}} \cdot \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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{y ^{2}} (u - ln ∣ u ∣)

Setze in

u = 1 + y x

ein

= \frac{1}{y ^{2}} (1 + y x - ln ∣ 1 + y x ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{y ^{2}} (1 + y x - ln ∣ 1 + y x ∣) + C

\frac{\partial}{\partial x} (\frac{3}{t + 1})

d er i v a t i v e f (t)

d er i v a t i v e f (x) = (1 + cos^{2} (x))^{7}

\frac{\partial ^{2}}{\partial u ( v ) \partial v} (8 v e^{- 6 u (v)})

\int \frac{2 cos ( x )}{- 2 + 6 sin ( x )} d x