integral of (x^3)/((x^2+a^2)^3)

Lösung

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

\frac{- 2 x ^{2} - a ^{2}}{4 ( x ^{2} + a ^{2} ) ^{2}} + C

Schritte zur Lösung

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

= a^{4} \cdot \int \frac{u ^{3}}{( a ^{2} u ^{2} + a ^{2} ) ^{3}} d u

Wende U-Substitution an

= a^{4} \cdot \int \frac{v - a ^{2}}{2 a ^{4} v ^{3}} d v

Entferne die Konstante:

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

= a^{4} \frac{1}{2 a ^{4}} \cdot \int \frac{v - a ^{2}}{v ^{3}} d v

Multipliziere aus

\frac{v - a ^{2}}{v ^{3}} : \frac{1}{v ^{2}} - \frac{a ^{2}}{v ^{3}}

= a^{4} \frac{1}{2 a ^{4}} \cdot \int \frac{1}{v ^{2}} - \frac{a ^{2}}{v ^{3}} 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

= a^{4} \frac{1}{2 a ^{4}} (\int \frac{1}{v ^{2}} d v - \int \frac{a ^{2}}{v ^{3}} d v)

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

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

= a^{4} \frac{1}{2 a ^{4}} (- \frac{1}{v} - (- \frac{a ^{2}}{2 v ^{2}}))

Ersetze zurück

= a^{4} \frac{1}{2 a ^{4}} - \frac{1}{a ^{2} ( \frac{x}{a} ) ^{2} + a ^{2}} - - \frac{a ^{2}}{2 ( a ^{2} ( \frac{x}{a} ) ^{2} + a ^{2} ) ^{2}}

Vereinfache

a^{4} \frac{1}{2 a ^{4}} - \frac{1}{a ^{2} ( \frac{x}{a} ) ^{2} + a ^{2}} - - \frac{a ^{2}}{2 ( a ^{2} ( \frac{x}{a} ) ^{2} + a ^{2} ) ^{2}} : \frac{- 2 x ^{2} - a ^{2}}{4 ( x ^{2} + a ^{2} ) ^{2}}

= \frac{- 2 x ^{2} - a ^{2}}{4 ( x ^{2} + a ^{2} ) ^{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{- 2 x ^{2} - a ^{2}}{4 ( x ^{2} + a ^{2} ) ^{2}} + C

a re a y = 3 x, x + y = 4, y = 1

\int_{- 1}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + x y + y ^{2}})

y^{^{''}} - 2 y^{^{'}} - 3 y = 9

a re a 5, x^{2} + 1