integral of (2x+3)/(4x^2-9)

Lösung

\int \frac{2 x + 3}{4 x ^{2} - 9} d x

\frac{1}{4} ln 4 x^{2} - 9 - \frac{1}{4} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1) + C

Schritte zur Lösung

\int \frac{2 x + 3}{4 x ^{2} - 9} d x

Multipliziere aus

\frac{2 x + 3}{4 x ^{2} - 9} : \frac{2 x}{4 x ^{2} - 9} + \frac{3}{4 x ^{2} - 9}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x}{4 x ^{2} - 9} d x + \int \frac{3}{4 x ^{2} - 9} d x

\int \frac{2 x}{4 x ^{2} - 9} d x = \frac{1}{4} ln 4 x^{2} - 9

\int \frac{3}{4 x ^{2} - 9} d x = - \frac{1}{4} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1)

= \frac{1}{4} ln 4 x^{2} - 9 - \frac{1}{4} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln 4 x^{2} - 9 - \frac{1}{4} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1) + C

x \to 0 lim (\frac{x + 2}{x ^{2}})

x \to - 4 lim (x^{2} + 3 x - 7)

\frac{\partial}{\partial y} (x z cos (x + yz))

\frac{d}{d x} (sin^{2} (ln (x)))

d er i v a t i v e \frac{3 x}{9 x ^{2} + 4 x - 9}