integral of ((x^2-3))/(x(x^4-4)^{1/2)}

Lösung

\int \frac{( x ^{2} - 3 )}{x ( x ^{4} - 4 ) ^{\frac{1}{2}}} d x

\frac{1}{4} (2 ln (\frac{1}{2} (x^{4} - 4)^{\frac{1}{2}} + x^{2}) - 3 arcsec (\frac{1}{2} x^{2})) + C

Schritte zur Lösung

\int \frac{x ^{2} - 3}{x ( x ^{4} - 4 ) ^{\frac{1}{2}}} d x

Wende U-Substitution an

= \int \frac{u - 3}{2 u ( u ^{2} - 4 ) ^{\frac{1}{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{u - 3}{u ( u ^{2} - 4 ) ^{\frac{1}{2}}} d u

Trigonometrische Substitution anwenden

= \frac{1}{2} \cdot \int \frac{2 sec ( v ) - 3}{2} d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 2 sec (v) - 3 d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot \frac{1}{2} (\int 2 sec (v) d v - \int 3 d v)

\int 2 sec (v) d v = 2 ln ∣ tan (v) + sec (v) ∣

\int 3 d v = 3 v

= \frac{1}{2} \cdot \frac{1}{2} (2 ln ∣ tan (v) + sec (v) ∣ - 3 v)

Ersetze zurück

= \frac{1}{2} \cdot \frac{1}{2} (2 ln tan (arcsec (\frac{1}{2} x^{2})) + sec (arcsec (\frac{1}{2} x^{2})) - 3 arcsec (\frac{1}{2} x^{2}))

Vereinfache

\frac{1}{2} \cdot \frac{1}{2} (2 ln tan (arcsec (\frac{1}{2} x^{2})) + sec (arcsec (\frac{1}{2} x^{2})) - 3 arcsec (\frac{1}{2} x^{2})) : \frac{1}{4} (2 ln (\frac{1}{2} (x^{4} - 4)^{\frac{1}{2}} + x^{2}) - 3 arcsec (\frac{1}{2} x^{2}))

= \frac{1}{4} (2 ln (\frac{1}{2} (x^{4} - 4)^{\frac{1}{2}} + x^{2}) - 3 arcsec (\frac{1}{2} x^{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (2 ln (\frac{1}{2} (x^{4} - 4)^{\frac{1}{2}} + x^{2}) - 3 arcsec (\frac{1}{2} x^{2})) + C

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