integral of (x-1)/((x+1)^3)

Lösung

\int \frac{x - 1}{( x + 1 ) ^{3}} d x

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

Schritte zur Lösung

\int \frac{x - 1}{( x + 1 ) ^{3}} d x

Wende U-Substitution an

= \int \frac{u - 2}{u ^{3}} d u

Multipliziere aus

\frac{u - 2}{u ^{3}} : \frac{1}{u ^{2}} - \frac{2}{u ^{3}}

= \int \frac{1}{u ^{2}} - \frac{2}{u ^{3}} 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 ^{2}} d u - \int \frac{2}{u ^{3}} d u

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

\int \frac{2}{u ^{3}} d u = - \frac{1}{u ^{2}}

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

Setze in

u = x + 1

ein

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

x \to 5 lim (x + 3)

y^{^{'}} + 1 = 2 y

\int \frac{x}{( x ^{2} - 4 ) ^{3}} d x

\frac{d}{d x} (\frac{a x ^{2}}{ln ( \frac{x}{4} )})

y^{^{'}} = \frac{x - y}{x}