y = x 2 y = 2 ( x 2 + 1 ) x = − 1 x = 1 {\displaystyle {\begin{aligned}&\color {red}\mathbf {y=x^{2}} &\color {royalblue}\mathbf {y={\frac {2}{({x^{2}}+1)}}} \\&x=-1&x=1\end{aligned}}}
∫ 1 − 1 ( 2 ( x 2 + 1 ) ) − ( x 2 ) d x = ∫ 1 − 1 ( 2 ⋅ 1 ( x 2 + 1 ) ) − ( x 2 ) d x = [ 2 arctan ( x ) − x 3 3 ] | − 1 1 = [ 2 arctan ( 1 ) − ( 1 ) 3 3 ] − [ ( 2 arctan ( − 1 ) − ( − 1 ) 3 3 ] = [ 2 π 4 − 1 3 ] − [ − 2 π 4 − ( − 1 3 ) ] = π 2 − 1 3 + π 2 − 1 3 = π − 2 3 {\displaystyle {\begin{aligned}\int _{1}^{-1}\left({\frac {2}{({x^{2}}+1)}}\right)-\left(x^{2}\right)dx&=\int _{1}^{-1}\left(2\cdot {\frac {1}{(x^{2}+1)}}\right)-\left(x^{2}\right)dx\\[2ex]&=\left[2\arctan(x)-{\frac {x^{3}}{3}}\right]{\Bigg |}_{-1}^{1}\\[2ex]&=\left[2\arctan(1)-{\frac {(1)^{3}}{3}}\right]-\left[(2\arctan(-1)-{\frac {(-1)^{3}}{3}}\right]\\[2ex]&=\left[{\frac {2\pi }{4}}-{\frac {1}{3}}\right]-\left[-{\frac {2\pi }{4}}-(-{\frac {1}{3}})\right]\\[2ex]&={\frac {\pi }{2}}-{\frac {1}{3}}+{\frac {\pi }{2}}-{\frac {1}{3}}\\[2ex]&=\pi -{\frac {2}{3}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{1}^{-1}\left({\frac {2}{({x^{2}}+1)}}\right)-\left(x^{2}\right)dx&=\int _{1}^{-1}\left(2\cdot {\frac {1}{(x^{2}+1)}}\right)-\left(x^{2}\right)dx\\[2ex]&=\left[2\arctan(x)-{\frac {x^{3}}{3}}\right]{\Bigg |}_{-1}^{1}\\[2ex]&=\left[2\arctan(1)-{\frac {(1)^{3}}{3}}\right]-\left[(2\arctan(-1)-{\frac {(-1)^{3}}{3}}\right]\\[2ex]&=\left[{\frac {2\pi }{4}}-{\frac {1}{3}}\right]-\left[-{\frac {2\pi }{4}}-(-{\frac {1}{3}})\right]\\[2ex]&={\frac {\pi }{2}}-{\frac {1}{3}}+{\frac {\pi }{2}}-{\frac {1}{3}}\\[2ex]&=\pi -{\frac {2}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/30b644a116db3c8f16a473bc7b279dcadb9d7a61)
