integral of 5/(x(x^4+4))

Lösung

\int \frac{5}{x ( x ^{4} + 4 )} d x

\frac{5}{4} (ln (x) - \frac{1}{4} ln x^{4} + 4) + C

Schritte zur Lösung

\int \frac{5}{x ( x ^{4} + 4 )} d x

Entferne die Konstante:

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

= 5 \cdot \int \frac{1}{x ( x ^{4} + 4 )} d x

Wende U-Substitution an

= 5 \cdot \int \frac{1}{4 u ( u + 4 )} d u

Entferne die Konstante:

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

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 4 )} d u

Ermittle den Partialbruch von

\frac{1}{u ( u + 4 )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u + 4 )} 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

= 5 \cdot \frac{1}{4} (\int \frac{1}{4 u} d u - \int \frac{1}{4 ( u + 4 )} d u)

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u + 4 )} d u = \frac{1}{4} ln ∣ u + 4 ∣

= 5 \cdot \frac{1}{4} (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 4 ∣)

Setze in

u = x^{4}

ein

= 5 \cdot \frac{1}{4} (\frac{1}{4} ln x^{4} - \frac{1}{4} ln x^{4} + 4)

Vereinfache

5 \cdot \frac{1}{4} (\frac{1}{4} ln x^{4} - \frac{1}{4} ln x^{4} + 4) : \frac{5}{4} (ln (x) - \frac{1}{4} ln x^{4} + 4)

= \frac{5}{4} (ln (x) - \frac{1}{4} ln x^{4} + 4)

Füge eine Konstante zur Lösung hinzu

= \frac{5}{4} (ln (x) - \frac{1}{4} ln x^{4} + 4) + C

\frac{\partial}{\partial x} (x y e^{- 8 (x^{2} + y^{2})})

\int_{0}^{1} 8 cos (2 x) d x

\int_{0}^{9} \frac{1}{9 - x} d x

\frac{d}{d x} (4 x^{\frac{5}{4}} + 10 x^{\frac{3}{2}} + 10 x)

\frac{d}{d x} ((x - 4) \cdot a x^{2} + ln (\frac{x}{4}) \cdot b x^{4})