What is the integral of arctan(1/(2x)) ?

The integral of arctan(1/(2x)) is xarctan(1/(2x))+1/4 ln|1+4x^2|+C

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integral of arctan(1/(2x))

integral

arctan (\frac{1}{2 x})

x arctan (\frac{1}{2 x}) + \frac{1}{4} ln 1 + 4 x^{2} + C

Solution steps

\int arctan (\frac{1}{2 x}) d x

Apply Integration By Parts

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

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

= x arctan (\frac{1}{2 x}) - (- \frac{1}{4} ln 1 + 4 x^{2})

Simplify

= x arctan (\frac{1}{2 x}) + \frac{1}{4} ln 1 + 4 x^{2}

Add a constant to the solution

= x arctan (\frac{1}{2 x}) + \frac{1}{4} ln 1 + 4 x^{2} + C

What is the integral of arctan(1/(2x)) ?
The integral of arctan(1/(2x)) is xarctan(1/(2x))+1/4 ln|1+4x^2|+C