integral of x/((1+x)^2)

Lösung

\int \frac{x}{( 1 + x ) ^{2}} d x

ln ∣ 1 + x ∣ + \frac{1}{1 + x} + C

Schritte zur Lösung

\int \frac{x}{( 1 + x ) ^{2}} d x

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 = 1 + x

ein

= ln ∣ 1 + x ∣ - (- \frac{1}{1 + x})

Vereinfache

= ln ∣ 1 + x ∣ + \frac{1}{1 + x}

Füge eine Konstante zur Lösung hinzu

= ln ∣ 1 + x ∣ + \frac{1}{1 + x} + C

x + y y^{^{'}} = 0, y (1) = 9

t an g e n t f (x) = 2 tan (x), at x = \frac{π}{4}

\frac{d}{d x} (2 x^{2} + 2)

\frac{\partial}{\partial y} ((1 + x y) e^{x y})

n = 1 \sum \infty \frac{9 ^{n}}{1 0 ^{n}}