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

Lösung

\int \frac{x}{4 + x ^{4}} d x

\frac{1}{2} ln (\frac{x ^{2} + 4 + x ^{4}}{2}) + C

Schritte zur Lösung

\int \frac{x}{4 + x ^{4}} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Trigonometrische Substitution anwenden

= \frac{1}{2} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= \frac{1}{2} ln tan (arctan (\frac{1}{2} x^{2})) + sec (arctan (\frac{1}{2} x^{2}))

Vereinfache

\frac{1}{2} ln tan (arctan (\frac{1}{2} x^{2})) + sec (arctan (\frac{1}{2} x^{2})) : \frac{1}{2} ln (\frac{x ^{2} + 4 + x ^{4}}{2})

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln (\frac{x ^{2} + 4 + x ^{4}}{2}) + C

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