integral of (-8x+162)/(x^2+9)

Lösung

\int \frac{- 8 x + 162}{x ^{2} + 9} d x

- 4 ln x^{2} + 9 + 54 arctan (\frac{x}{3}) + C

Schritte zur Lösung

\int \frac{- 8 x + 162}{x ^{2} + 9} d x

Multipliziere aus

\frac{- 8 x + 162}{x ^{2} + 9} : - \frac{8 x}{x ^{2} + 9} + \frac{162}{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{8 x}{x ^{2} + 9} d x + \int \frac{162}{x ^{2} + 9} d x

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

\int \frac{162}{x ^{2} + 9} d x = 54 arctan (\frac{x}{3})

= - 4 ln x^{2} + 9 + 54 arctan (\frac{x}{3})

Füge eine Konstante zur Lösung hinzu

= - 4 ln x^{2} + 9 + 54 arctan (\frac{x}{3}) + C

x \frac{d y}{d x} = x^{2} + 2 x - 3

\int e^{x} e^{e^{x}} d x

n = 2 \sum \infty \frac{1}{n ^{2} + 6 n + 8}

t a y l or - (3 x - 4)^{3}, 3

\int_{- 1}^{64} \frac{1}{x ^{\frac{1}{3}}} d x