English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x/(sqrt(x^4-12x^2+35))

Lösung

\int \frac{x}{x ^{4} - 12 x ^{2} + 35} 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} - 12 x ^{2} + 35} d x

Wende U-Substitution an

= \int \frac{1}{4 u u + 35 - 12 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 + 35 - 12 u} d u

Wende Exponentenregel an:

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

\frac{1}{u u + 35 - 12 u} = u^{- \frac{1}{2}} \frac{1}{u + 35 - 12}

= \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{u + 35 - 12} d u

Vereinfache

u^{- \frac{1}{2}} \frac{1}{u + 35 - 12} : \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

\int \frac{x ^{2}}{( 1 + x ^{3} ) ^{1.9}} d x

\int 3 (x + 1) 2 d x

\int \frac{e ^{6 x}}{e ^{6 x} + 6} d x

\int (sec (3 u) + csc (3 u))^{2} d u

\int \frac{- e ^{2}}{x} + 3 \cdot 3^{x} d x