integral of 1/(sqrt(7+8x^2))

Lösung

\int \frac{1}{7 + 8 x ^{2}} d x

\frac{1}{2 2} ln \frac{2 2}{7} x + \frac{1}{7} (7 + 8 x^{2}) + C

Schritte zur Lösung

\int \frac{1}{7 + 8 x ^{2}} d x

Trigonometrische Substitution anwenden

= \int \frac{sec ( u )}{2 2} d u

Entferne die Konstante:

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

= \frac{1}{2 2} \cdot \int sec (u) d u

Nutze das gemeinsame Integral :

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

= \frac{1}{2 2} ln ∣ tan (u) + sec (u) ∣

Setze in

u = arctan (\frac{2 2}{7} x)

ein

= \frac{1}{2 2} ln tan (arctan (\frac{2 2}{7} x)) + sec (arctan (\frac{2 2}{7} x))

Vereinfache

\frac{1}{2 2} ln tan (arctan (\frac{2 2}{7} x)) + sec (arctan (\frac{2 2}{7} x)) : \frac{1}{2 2} ln \frac{2 2}{7} x + \frac{1}{7} (7 + 8 x^{2})

= \frac{1}{2 2} ln \frac{2 2}{7} x + \frac{1}{7} (7 + 8 x^{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2 2} ln \frac{2 2}{7} x + \frac{1}{7} (7 + 8 x^{2}) + C

n = 0 \sum \infty n! (x - 4)^{n}

\frac{d}{d x} (arctan (\frac{2 x}{1 + x ^{2}}))

\frac{d}{d t} (3 t + 6 t^{3})

t an g e n t \frac{x ^{2}}{x - 1}

d er i v a t i v e f (x) = \frac{1}{ln ( x )}