We will use substitution for the partial integration:\(u=arc tan ( \frac{1}{x}), dv = dx \\ du= \frac{-dx}{(1+ x^{2}) x^{2} } , v=x\)
The integral becomes:=\(x arc tan x- \int {x \frac{-dx}{(1+ x^{2}) x^{2} } } \, dx = \\ =x arc tan x+ \int { \frac{dx}{(1+ x^{2} )x} } \, dx\)
2nd integration:
we will add and subtract x^2 to the numerator:\(\int { \frac{1+ x^{2} - x^{2} }{(1+ x^{2} )x} } \, dx= \\ \int { \frac{1}{x} } \, dx- \int { \frac{x}{1+ x^{2} } } \, dx = \\ ln(x) - \int { \frac{x}{1+ x^{2} } } \, dx\)
u-substitution:\(u=1+ x^{2} , du=2xdx, xdx= \frac{du}{2}\)
...=\(ln(x)- \frac{1}{2} \int { \frac{1}{u} } \, du=ln(x)- \frac{1}{2} ln(u)= \\ ln(x)-ln( \sqrt{1+ x^{2} )} =ln \frac{x}{ \sqrt{1+ x^{2} } }\)
Finally:...=\(xarctan( \frac{1}{x})+ln( \frac{x}{ \sqrt{1+ x^{2} } })+C\)
