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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

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

= \int u^{2} - 3 u + 3 - \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

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

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

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

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

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

Setze in

u = 1 + x

ein

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

Füge eine Konstante zur Lösung hinzu

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

y^{^{''}} - y^{^{'}} - 2 y = - 6 t + 4 t^{2}

\int_{- 2}^{4} ∣ - 2 x + 4 ∣ d x

y^{^{'}} - \frac{2}{x} y = 0

\frac{\partial}{\partial x} (\frac{7 x ^{2}}{( x - y ) ^{2}})

\int_{1}^{t} \frac{3}{x ^{3}} d x