What is the integral of arccot(4y) ?

The integral of arccot(4y) is yarccot(4y)+1/8 ln|16y^2+1|+C

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integral of arccot(4y)

\int arccot (4 y) d y

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

Solution steps

\int arccot (4 y) d y

Apply Integration By Parts

: y arccot (4 y) - \int - \frac{4 y}{16 y ^{2} + 1} d y

= y arccot (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

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

Simplify

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

Add a constant to the solution

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

x \to x^{1} lim (x^{2})

y = 3 x - 7

\int_{0}^{1} (x^{2} - 2 x + 3) d x

\int_{0}^{\frac{1}{2}} 7 arccos (x) d x

\frac{\partial}{\partial y} (1 - x y cos (π y))

What is the integral of arccot(4y) ?
The integral of arccot(4y) is yarccot(4y)+1/8 ln|16y^2+1|+C