What is the integral of arccot(4x) ?

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

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integral of arccot(4x)

\int arccot (4 x) d x

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

Solution steps

\int arccot (4 x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int (3 x + 10)^{12} d x

\int (1 + y^{2})^{- \frac{3}{2}} d y

\int_{- \infty}^{3} \frac{3}{x ^{2} + 9} d x

f (x) = x + 2, x = 1, x = 3

y^{^{'}} = \frac{x - y}{x + 2 y}

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