∫ 1 3 a r c t a n ( 1 x ) d x {\displaystyle \int _{1}^{\sqrt {3}}arctan\left({\frac {1}{x}}\right)dx}
l e t u = a r c t a n ( 1 x ) d u = d x = d u = 1 1 + ( 1 / x ) 2 x − 1 x 2 d x = − d x x 2 + 1 {\displaystyle 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}}}
∫ 1 3 a r c t a n 1 x d x = [ x a r c t a n 1 x ] | 0 1 + ∫ 1 3 x d x x 2 + 1 = 3 π 6 = 1 4 = 1 2 [ i n ( x 2 + 1 ) ] | 1 3 {\displaystyle \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}}}
= π 3 6 − π 4 ( i n 4 − i n 2 ) = π 3 6 − π 2 = 1 2 i n 4 2 = π 3 6 − π 2 + 1 2 i n 2 {\displaystyle ={\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}
![{\displaystyle \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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6d041815f57135b8774b91144a0ac47a8028fd1f)
