integral of (200)/(x(100+x^4))

Lösung

\int \frac{200}{x ( 100 + x ^{4} )} d x

50 (\frac{1}{25} ln (x) - \frac{1}{100} ln x^{4} + 100) + C

Schritte zur Lösung

\int \frac{200}{x ( 100 + x ^{4} )} d x

Entferne die Konstante:

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

= 200 \cdot \int \frac{1}{x ( 100 + x ^{4} )} d x

Wende U-Substitution an

= 200 \cdot \int \frac{1}{4 u ( 100 + u )} d u

Entferne die Konstante:

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

= 200 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( 100 + u )} d u

Ermittle den Partialbruch von

\frac{1}{u ( 100 + u )} : \frac{1}{100 u} - \frac{1}{100 ( u + 100 )}

= 200 \cdot \frac{1}{4} \cdot \int \frac{1}{100 u} - \frac{1}{100 ( u + 100 )} 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

= 200 \cdot \frac{1}{4} (\int \frac{1}{100 u} d u - \int \frac{1}{100 ( u + 100 )} d u)

\int \frac{1}{100 u} d u = \frac{1}{100} ln ∣ u ∣

\int \frac{1}{100 ( u + 100 )} d u = \frac{1}{100} ln ∣ u + 100 ∣

= 200 \cdot \frac{1}{4} (\frac{1}{100} ln ∣ u ∣ - \frac{1}{100} ln ∣ u + 100 ∣)

Setze in

u = x^{4}

ein

= 200 \cdot \frac{1}{4} (\frac{1}{100} ln x^{4} - \frac{1}{100} ln x^{4} + 100)

Vereinfache

200 \cdot \frac{1}{4} (\frac{1}{100} ln x^{4} - \frac{1}{100} ln x^{4} + 100) : 50 (\frac{1}{25} ln (x) - \frac{1}{100} ln x^{4} + 100)

= 50 (\frac{1}{25} ln (x) - \frac{1}{100} ln x^{4} + 100)

Füge eine Konstante zur Lösung hinzu

= 50 (\frac{1}{25} ln (x) - \frac{1}{100} ln x^{4} + 100) + C

\frac{\partial}{\partial x} (e^{x} sin (10 y + 5 z))

\int (3 x + 4) d x

\int \frac{( 6 x - 1 )}{x ^{3} ( 2 x - 1 )} d x

d er i v a t i v e 4 a x - x

f^{^{'}} (x) = 0