integral of 1/(sqrt(4x^2+8x+9))

Lösung

\int \frac{1}{4 x ^{2} + 8 x + 9} d x

\frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{4 x ^{2} + 8 x + 9}{5} + C

Schritte zur Lösung

\int \frac{1}{4 x ^{2} + 8 x + 9} d x

Vervollständige das Quadrat

4 x^{2} + 8 x + 9 : 4 (x + 1)^{2} + 5

= \int \frac{1}{4 ( x + 1 ) ^{2} + 5} d x

Wende U-Substitution an

= \int \frac{1}{4 u ^{2} + 5} d u

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Ersetze zurück

= \frac{1}{2} ln tan (arctan (\frac{2}{5} (x + 1))) + sec (arctan (\frac{2}{5} (x + 1)))

Vereinfache

\frac{1}{2} ln tan (arctan (\frac{2}{5} (x + 1))) + sec (arctan (\frac{2}{5} (x + 1))) : \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{4 x ^{2} + 8 x + 9}{5}

= \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{4 x ^{2} + 8 x + 9}{5}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln \frac{2}{5} (x + 1) + \frac{4 x ^{2} + 8 x + 9}{5} + C

\int 6 - 2 x d x

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

\int \frac{3 x}{e ^{2 x}} d x

\int \frac{1 + sin ^{2} ( x )}{1 + cos ( 2 x )} d x

\int x^{4} + 7 x d x