integral of 6arctan(4y)

Lösung

\int 6 arctan (4 y) d y

6 (y arctan (4 y) - \frac{1}{8} ln 16 y^{2} + 1) + C

Schritte zur Lösung

\int 6 arctan (4 y) d y

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int arctan (4 y) d y

Wende die partielle Integration an

= 6 (y arctan (4 y) - \int \frac{4 y}{16 y ^{2} + 1} d y)

\int \frac{4 y}{16 y ^{2} + 1} d y = \frac{1}{8} ln 16 y^{2} + 1

= 6 (y arctan (4 y) - \frac{1}{8} ln 16 y^{2} + 1)

Füge eine Konstante zur Lösung hinzu

= 6 (y arctan (4 y) - \frac{1}{8} ln 16 y^{2} + 1) + C

\frac{\partial}{\partial x} (2 y^{4 x})

t an g e n t f (x) = sin (x) + 1, at x = - \frac{π}{2}

a re a f (x) = x^{2} - 4, x = - 4, x = 3

\int \frac{1}{x ^{2} a ^{2} - x ^{2}} d x

x \to - 3 lim (∣ x ∣ - 2)