integral of 1/(sqrt(64+x^2))

Lösung

\int \frac{1}{64 + x ^{2}} d x

ln (\frac{1}{8} x + 64 + x^{2}) + C

Schritte zur Lösung

\int \frac{1}{64 + x ^{2}} d x

Trigonometrische Substitution anwenden

= \int sec (u) d u

Nutze das gemeinsame Integral :

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

= ln ∣ tan (u) + sec (u) ∣

Setze in

u = arctan (\frac{1}{8} x)

ein

= ln tan (arctan (\frac{1}{8} x)) + sec (arctan (\frac{1}{8} x))

Vereinfache

ln tan (arctan (\frac{1}{8} x)) + sec (arctan (\frac{1}{8} x)) : ln (\frac{1}{8} x + 64 + x^{2})

= ln (\frac{1}{8} x + 64 + x^{2})

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{8} x + 64 + x^{2}) + C

f (x) = x cos (x)

\frac{d}{d x} (ln (x) + 2 x)

t a y l or \frac{1}{1 + x ^{2}}, 0

\int_{0}^{1} (1 - x^{7})^{2} d x

d er i v a t i v e x^{3} - 2 x^{2} - 4 x + 6