integral of 1/(sqrt(5x^2-x-1))

Lösung

\int \frac{1}{5 x ^{2} - x - 1} d x

\frac{1}{5} ln \frac{100 x ^{2} - 20 x - 20}{21} + \frac{10}{21} x - \frac{1}{21} + C

Schritte zur Lösung

\int \frac{1}{5 x ^{2} - x - 1} d x

Vervollständige das Quadrat

5 x^{2} - x - 1 : 5 (x - \frac{1}{10})^{2} - \frac{21}{20}

= \int \frac{1}{5 ( x - \frac{1}{10} ) ^{2} - \frac{21}{20}} d x

Wende U-Substitution an

= \int \frac{2 5}{100 u ^{2} - 21} d u

Entferne die Konstante:

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

= 25 \cdot \int \frac{1}{100 u ^{2} - 21} d u

Trigonometrische Substitution anwenden

= 25 \cdot \int \frac{sec ( v )}{10} d v

Entferne die Konstante:

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

= 25 \frac{1}{10} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= 25 \frac{1}{10} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= 25 \frac{1}{10} ln tan (arcsec (\frac{10}{21} (x - \frac{1}{10}))) + sec (arcsec (\frac{10}{21} (x - \frac{1}{10})))

Vereinfache

25 \frac{1}{10} ln tan (arcsec (\frac{10}{21} (x - \frac{1}{10}))) + sec (arcsec (\frac{10}{21} (x - \frac{1}{10}))) : \frac{1}{5} ln \frac{100 x ^{2} - 20 x - 20}{21} + \frac{10}{21} x - \frac{1}{21}

= \frac{1}{5} ln \frac{100 x ^{2} - 20 x - 20}{21} + \frac{10}{21} x - \frac{1}{21}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} ln \frac{100 x ^{2} - 20 x - 20}{21} + \frac{10}{21} x - \frac{1}{21} + C

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\int sec^{13} (x) tan (x) d x