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

The integral of (arctan(1/x))/(1+x^2) is arctan(1/x)arctan(x)+1/2 arctan^2(x)+C

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integral of (arctan(1/x))/(1+x^2)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

arctan (\frac{1}{x}) arctan (x) - (- \frac{arctan ^{2} ( x )}{2}) : arctan (\frac{1}{x}) arctan (x) + \frac{1}{2} arctan^{2} (x)

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

Add a constant to the solution

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

\int 18 x d x

\int e^{2 x} arctan (e^{x} - 1) d x

\int sin^{2} (x) tan (x) d x

\int \frac{1}{x ^{2} - 8 x + 25} d x

x^{- 2}

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