integral of 1/(sqrt(x^2+6x+25))

Lösung

\int \frac{1}{x ^{2} + 6 x + 25} d x

ln (\frac{1}{4} x + 3 + x^{2} + 6 x + 25) + C

Schritte zur Lösung

\int \frac{1}{x ^{2} + 6 x + 25} d x

Vervollständige das Quadrat

x^{2} + 6 x + 25 : (x + 3)^{2} + 16

= \int \frac{1}{( x + 3 ) ^{2} + 16} d x

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} (x + 3))) + sec (arctan (\frac{1}{4} (x + 3)))

Vereinfache

ln tan (arctan (\frac{1}{4} (x + 3))) + sec (arctan (\frac{1}{4} (x + 3))) : ln (\frac{1}{4} x + 3 + x^{2} + 6 x + 25)

= ln (\frac{1}{4} x + 3 + x^{2} + 6 x + 25)

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{4} x + 3 + x^{2} + 6 x + 25) + C

ma c l a u r in \frac{1}{x + 1}

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

\int sin^{2} (\frac{π x}{L}) d x

t^{2} y^{^{''}} - 3 t y^{^{'}} + 4 y = 0

\frac{\partial}{\partial y} ((x + y)^{5} + (x - y)^{5})