integral of (1+1/u)^{-3}(1/(u^2))

Lösung

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

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

Schritte zur Lösung

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

Wende die partielle Integration an

= - \frac{u ^{2}}{( u + 1 ) ^{3}} - \int - \frac{3 u}{( u + 1 ) ^{4}} d u

\int - \frac{3 u}{( u + 1 ) ^{4}} d u = \frac{3}{2 ( u + 1 ) ^{2}} - \frac{1}{( u + 1 ) ^{3}}

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (4 x + 8 x^{\frac{3}{8}})

\frac{d}{d x} \frac{( x ^{2} - 4 ) ^{3}}{36}

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

s l o p e in t erce pt (- 5, - 1), (0, 3)

\frac{d y}{d x} = \frac{y + 1}{x ^{3}}