integral of arccot(5y)

Lösung

\int arccot (5 y) d y

y arccot (5 y) + \frac{1}{10} ln 25 y^{2} + 1 + C

Schritte zur Lösung

\int arccot (5 y) d y

Wende die partielle Integration an

= y arccot (5 y) - \int - \frac{5 y}{25 y ^{2} + 1} d y

\int - \frac{5 y}{25 y ^{2} + 1} d y = - \frac{1}{10} ln 25 y^{2} + 1

= y arccot (5 y) - (- \frac{1}{10} ln 25 y^{2} + 1)

Vereinfache

= y arccot (5 y) + \frac{1}{10} ln 25 y^{2} + 1

Füge eine Konstante zur Lösung hinzu

= y arccot (5 y) + \frac{1}{10} ln 25 y^{2} + 1 + C

t an g e n t f (x) = x^{2} - 6, (4, 10)

\frac{d ^{2}}{d x ^{2}} (x e^{4 x})

t an g e n t f (x) = \frac{1}{2 x}, at x = 81

\frac{d}{d x} (\frac{1}{( x ^{3} + 4 ) ^{2}})

x \to 0 - lim (x ln (x))