integral of (x^7)/((1+x^4)^2)

Lösung

\int \frac{x ^{7}}{( 1 + x ^{4} ) ^{2}} d x

\frac{1}{4} (ln 1 + x^{4} + \frac{1}{1 + x ^{4}}) + C

Schritte zur Lösung

\int \frac{x ^{7}}{( 1 + x ^{4} ) ^{2}} d x

Wende U-Substitution an

= \int \frac{u - 1}{4 u ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int \frac{u - 1}{u ^{2}} d u

Multipliziere aus

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

= \frac{1}{4} \cdot \int \frac{1}{u} - \frac{1}{u ^{2}} 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= \frac{1}{4} (ln ∣ u ∣ - (- \frac{1}{u}))

Setze in

u = 1 + x^{4}

ein

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

Vereinfache

= \frac{1}{4} (ln 1 + x^{4} + \frac{1}{1 + x ^{4}})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (ln 1 + x^{4} + \frac{1}{1 + x ^{4}}) + C

d er i v a t i v e (\frac{( e ^{6} )}{( e ^{4} )})^{- 1}

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 4})

x \frac{d v}{d x} = \frac{1 - 4 v ^{2}}{3 v}

\frac{d}{d x} (lo g_{3} (\frac{3 x ^{2}}{x}))

\frac{d y}{d x} = \frac{x y + x + 3 y + 3}{x y + 2 x - y - 2}