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

Lösung

integral

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

\frac{( u + 1 ) ^{3}}{3} - \frac{3 ( u + 1 ) ^{2}}{2} + 3 (u + 1) - ln ∣ u + 1 ∣ + C

Schritte zur Lösung

\int \frac{u ^{3}}{u + 1} d u

Wende U-Substitution an

= \int \frac{( v - 1 ) ^{3}}{v} d v

Multipliziere aus

\frac{( v - 1 ) ^{3}}{v} : v^{2} - 3 v + 3 - \frac{1}{v}

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int v^{2} d v = \frac{v ^{3}}{3}

\int 3 v d v = \frac{3 v ^{2}}{2}

\int 3 d v = 3 v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{v ^{3}}{3} - \frac{3 v ^{2}}{2} + 3 v - ln ∣ v ∣

Setze in

v = u + 1

ein

= \frac{( u + 1 ) ^{3}}{3} - \frac{3 ( u + 1 ) ^{2}}{2} + 3 (u + 1) - ln ∣ u + 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{( u + 1 ) ^{3}}{3} - \frac{3 ( u + 1 ) ^{2}}{2} + 3 (u + 1) - ln ∣ u + 1 ∣ + C

in t e g r a l 1 - y^{2}

in t e g r a l x^{5} (x^{2} + 3)^{4}

in t e g r a l (4 + x)^{\frac{1}{2}}

in t e g r a l x^{2} + 2 x + 3

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