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

Lösung

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

- 6 arctan (\frac{x}{3}) + x + C

Schritte zur Lösung

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

\frac{x ^{2} - 9}{x ^{2} + 9} = - \frac{18}{x ^{2} + 9} + 1

= \int - \frac{18}{x ^{2} + 9} + 1 d x

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{18}{x ^{2} + 9} d x + \int 1 d x

\int \frac{18}{x ^{2} + 9} d x = 6 arctan (\frac{x}{3})

\int 1 d x = x

= - 6 arctan (\frac{x}{3}) + x

Füge eine Konstante zur Lösung hinzu

= - 6 arctan (\frac{x}{3}) + x + C

\int 3 v d v

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

\int cot^{5} (x) csc^{6} (x) d x

\int e^{3 y} (1 - e^{3 y})^{\frac{3}{2}} d y

\int - 3^{4} x d x