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

Lösung

integral

\frac{x}{( x + 1 ) ^{3}}

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

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

Setze in

u = x + 1

ein

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l y = x^{3} (e^{3})^{4}

in t e g r a l (e^{t})^{2}

in t e g r a l e^{\frac{1}{x}} x

in t e g r a l \frac{( x - 1 ) ^{2}}{x ^{2}}

in t e g r a l \frac{t}{6}