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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Ermittle den Partialbruch von

\frac{1}{( u - 1 ) u} : \frac{1}{u - 1} - \frac{1}{u}

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

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

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

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

Setze in

u = 1 + x^{4}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

s l o p e y = (3 x^{3} + 2 x)^{\frac{1}{2}}

d er i v a t i v e f (x) = e^{- 9}

\int_{0}^{0.08} 2000 x d x

\frac{d}{d x} (\frac{2 x}{x + 2})

\int 2 + 2 sin (x) d x