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

Lösung

integral

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + ln ∣ u ∣

Setze in

u = 1 + x

ein

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

Füge eine Konstante zur Lösung hinzu

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

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

in t e g r a l \frac{2}{e ^{x}}

in t e g r a l cos (\frac{1}{x})

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

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