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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

\frac{1 + x}{4 - 9 x ^{2}} : \frac{1}{4 - 9 x ^{2}} + \frac{x}{4 - 9 x ^{2}}

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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