integral of 1/(sqrt(t^2-10t+41))

Lösung

\int \frac{1}{t ^{2} - 10 t + 41} d t

ln (\frac{1}{4} t - 5 + t^{2} - 10 t + 41) + C

Schritte zur Lösung

\int \frac{1}{t ^{2} - 10 t + 41} d t

Vervollständige das Quadrat

t^{2} - 10 t + 41 : (t - 5)^{2} + 16

= \int \frac{1}{( t - 5 ) ^{2} + 16} d t

Wende U-Substitution an

= \int \frac{1}{u ^{2} + 16} d u

Trigonometrische Substitution anwenden

= \int sec (v) d v

Nutze das gemeinsame Integral :

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

= ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= ln tan (arctan (\frac{1}{4} (t - 5))) + sec (arctan (\frac{1}{4} (t - 5)))

Vereinfache

ln tan (arctan (\frac{1}{4} (t - 5))) + sec (arctan (\frac{1}{4} (t - 5))) : ln (\frac{1}{4} t - 5 + t^{2} - 10 t + 41)

= ln (\frac{1}{4} t - 5 + t^{2} - 10 t + 41)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{4} t - 5 + t^{2} - 10 t + 41) + C

f (x) = x^{2} arctan (5 x)

\int \frac{x + 5}{( x + 2 ) ( x - 1 )} d x

t an g e n t y = 5 + 5 x^{2} - 2 x^{3}, (1, 8)

x \to \infty lim (\frac{2}{3 x})

\int 10000 (t - 6) d t