integral of (x^5)/((x^3+1)^2)

Lösung

\int \frac{x ^{5}}{( x ^{3} + 1 ) ^{2}} d x

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

Schritte zur Lösung

\int \frac{x ^{5}}{( x ^{3} + 1 ) ^{2}} d x

Wende U-Substitution an

= \int \frac{u ^{\frac{1}{5}}}{5 ( u ^{\frac{3}{5}} + 1 ) ^{2}} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int \frac{u ^{\frac{1}{5}}}{( u ^{\frac{3}{5}} + 1 ) ^{2}} d u

Wende U-Substitution an

= \frac{1}{5} \cdot \int \frac{5 ( v - 1 )}{3 v ^{2}} d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int \frac{v - 1}{v ^{2}} d v

Multipliziere aus

\frac{v - 1}{v ^{2}} : \frac{1}{v} - \frac{1}{v ^{2}}

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int \frac{1}{v} - \frac{1}{v ^{2}} 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

= \frac{1}{5} \cdot \frac{5}{3} (\int \frac{1}{v} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= \frac{1}{5} \cdot \frac{5}{3} (ln ∣ v ∣ - (- \frac{1}{v}))

Ersetze zurück

= \frac{1}{5} \cdot \frac{5}{3} (ln (x^{5})^{\frac{3}{5}} + 1 - (- \frac{1}{( x ^{5} ) ^{\frac{3}{5}} + 1}))

Vereinfache

\frac{1}{5} \cdot \frac{5}{3} (ln (x^{5})^{\frac{3}{5}} + 1 - (- \frac{1}{( x ^{5} ) ^{\frac{3}{5}} + 1})) : \frac{1}{3} (ln x^{3} + 1 + \frac{1}{x ^{3} + 1})

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

Füge eine Konstante zur Lösung hinzu

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

x \to 1 lim (cos (\frac{π}{2} x))

\frac{d}{d x} ((x - 5)^{\frac{2}{3}})

\frac{d}{d x} (arcsin (- 3 x))

y^{^{''}} = - 4 y - 4 y^{^{'}}

\frac{\partial}{\partial y} (tan (y))