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

Lösung

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

ln ∣ x ∣ - \frac{4}{3} arctan (3 x) + C

Schritte zur Lösung

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{4}{3} arctan (3 x)

Füge eine Konstante zur Lösung hinzu

= ln ∣ x ∣ - \frac{4}{3} arctan (3 x) + C

t an g e n t f (x) = \frac{3}{2 x + 1}, at x = 4

\int_{- π}^{π} x e^{s i n (x^{2})} d x

\frac{d}{d x} (4 x^{2} - 2 x + 1)

\int x^{2} - 4 cos (x) + 8 d x

\int (- 4 x^{4} + 5 sin (x)) d x