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integral of x/(sqrt(x^4-16x^2+39))

Lösung

\int \frac{x}{x ^{4} - 16 x ^{2} + 39} d x

\frac{1}{2} ln \frac{1}{23} x^{2} + \frac{1}{23} (23 + x^{4}) + C

Schritte zur Lösung

\int \frac{x}{x ^{4} - 16 x ^{2} + 39} d x

Wende U-Substitution an

= \int \frac{1}{4 u u + 39 - 16 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 u + 39 - 16 u} d u

Wende Exponentenregel an:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{u u + 39 - 16 u} = u^{- \frac{1}{2}} \frac{1}{u + 39 - 16}

= \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{u + 39 - 16} d u

Vereinfache

u^{- \frac{1}{2}} \frac{1}{u + 39 - 16} : \frac{1}{u u + 23}

= \frac{1}{4} \cdot \int \frac{1}{u u + 23} d u

Wende U-Substitution an

= \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 23} d v

Entferne die Konstante:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 23} d v

Trigonometrische Substitution anwenden

= \frac{1}{4} \cdot 2 \cdot \int sec (w) d w

Nutze das gemeinsame Integral :

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

= \frac{1}{4} \cdot 2 ln ∣ tan (w) + sec (w) ∣

Ersetze zurück

= \frac{1}{4} \cdot 2 ln tan (arctan (\frac{1}{23} x^{4})) + sec (arctan (\frac{1}{23} x^{4}))

Vereinfache

\frac{1}{4} \cdot 2 ln tan (arctan (\frac{1}{23} x^{4})) + sec (arctan (\frac{1}{23} x^{4})) : \frac{1}{2} ln \frac{1}{23} x^{2} + \frac{1}{23} (23 + x^{4})

= \frac{1}{2} ln \frac{1}{23} x^{2} + \frac{1}{23} (23 + x^{4})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln \frac{1}{23} x^{2} + \frac{1}{23} (23 + x^{4}) + C

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