integral of 1/(sqrt(x^2+3/4))

Lösung

\int \frac{1}{x ^{2} + \frac{3}{4}} d x

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

Schritte zur Lösung

\int \frac{1}{x ^{2} + \frac{3}{4}} d x

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

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

Vereinfache

\frac{1}{\frac{3}{4} ( \frac{x ^{2}}{\frac{3}{4}} + 1 )} : \frac{2}{4 x ^{2} + 3}

= \int \frac{2}{4 x ^{2} + 3} d x

Entferne die Konstante:

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

= 2 \cdot \int \frac{1}{4 x ^{2} + 3} d x

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Setze in

u = arctan (\frac{2}{3} x)

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int - \frac{60}{x ^{2}} d x

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