integral of 1/(sqrt(5+3x^2))

Lösung

\int \frac{1}{5 + 3 x ^{2}} d x

\frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 3 x^{2}) + C

Schritte zur Lösung

\int \frac{1}{5 + 3 x ^{2}} d x

Trigonometrische Substitution anwenden

= \int \frac{sec ( u )}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int sec (u) d u

Nutze das gemeinsame Integral :

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

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Setze in

u = arctan (\frac{3}{5} x)

ein

= \frac{1}{3} ln tan (arctan (\frac{3}{5} x)) + sec (arctan (\frac{3}{5} x))

Vereinfache

\frac{1}{3} ln tan (arctan (\frac{3}{5} x)) + sec (arctan (\frac{3}{5} x)) : \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 3 x^{2})

= \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 3 x^{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 3 x^{2}) + C

\frac{d y}{d x} = \frac{1}{2} x + 1

a re a f (x) = \frac{1}{x}, g (x) = \frac{1}{3}, [2, 5]

d er i v a t i v e \frac{7 u ^{2}}{( u ^{2} + u ) ^{3}}

\int \frac{3 x ^{2}}{1 - x ^{6}} d x

a re a x, x^{2} - 2