7.1 Integration By Parts/26

$\int _{1}^{\sqrt {3}}arctan\left({\frac {1}{x}}\right)dx$

$letu=arctan\left({\frac {1}{x}}\right)du=dx=du={\frac {1}{1+(1/x)^{2}}}x{\frac {-1}{x^{2}}}dx={\frac {-dx}{x^{2}+1}}$

$\int _{1}^{\sqrt {3}}arctan{\frac {1}{x}}dx=[xarctan{\frac {1}{x}}]{\bigg |}_{0}^{1}+\int _{1}^{\sqrt {3}}{\frac {x}{dx}}x^{2}+1={\sqrt {3}}{\frac {\pi }{6}}={\frac {1}{4}}={\frac {1}{2}}[in(x^{2}+1)]{\bigg |}_{1}^{\sqrt {3}}$

$={\frac {\pi {\sqrt {3}}}{6}}-{\frac {\pi }{4}}(in4-in2)={\frac {\pi {\sqrt {3}}}{6}}-{\frac {\pi }{2}}={\frac {1}{2}}in{\frac {4}{2}}={\frac {\pi {\sqrt {3}}}{6}}-{\frac {\pi }{2}}+{\frac {1}{2}}in2$