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Dalzell's integral
π = 22 7 − ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x {\displaystyle \pi ={\frac {22}{7}}-\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}}}dx}
π = 3 + 2 ∫ 0 1 x ( 1 − x ) 2 1 + x 2 d x {\displaystyle \pi =3+2\int _{0}^{1}{\frac {x(1-x)^{2}}{1+x^{2}}}dx}
π = 3 + 4 ∫ 0 1 x 4 ( 1 − x ) ( 2 + x ) 1 + x + x 2 + x 3 d x {\displaystyle \pi =3+4\int _{0}^{1}{\frac {x^{4}(1-x)(2+x)}{1+x+x^{2}+x^{3}}}dx}
π = 4 − 4 ∫ 0 1 x 2 1 + x 2 d x {\displaystyle \pi =4-4\int _{0}^{1}{\frac {x^{2}}{1+x^{2}}}dx}
π = 10 3 − ∫ 0 1 ( 1 − x ) 4 1 + x 2 d x {\displaystyle \pi ={\frac {10}{3}}-\int _{0}^{1}{\frac {(1-x)^{4}}{1+x^{2}}}dx}
π = 31 10 + 2 ∫ 0 1 x 2 ( 1 − x ) 2 ( 1 − x + x 2 ) 1 + x 2 d x {\displaystyle \pi ={\frac {31}{10}}+2\int _{0}^{1}{\frac {x^{2}(1-x)^{2}(1-x+x^{2})}{1+x^{2}}}dx}
π = 31 10 + 4 ∫ 0 1 x 8 ( 1 − x ) ( 2 + x ) 1 + x + x 2 + x 3 d x {\displaystyle \pi ={\frac {31}{10}}+4\int _{0}^{1}{\frac {x^{8}(1-x)(2+x)}{1+x+x^{2}+x^{3}}}dx}
π = 16 5 − ∫ 0 1 x 2 ( 1 − x ) 2 ( 1 + 2 x + x 2 ) 1 + x 2 d x {\displaystyle \pi ={\frac {16}{5}}-\int _{0}^{1}{\frac {x^{2}(1-x)^{2}(1+2x+x^{2})}{1+x^{2}}}dx}
π = 19 6 − 2 ∫ 0 1 x 3 ( 1 − x ) 2 1 + x 2 d x {\displaystyle \pi ={\frac {19}{6}}-2\int _{0}^{1}{\frac {x^{3}(1-x)^{2}}{1+x^{2}}}dx}
π = 25 8 + 1 4 ∫ 0 1 x ( 1 − x ) 4 ( 1 + 4 x + x 2 ) 1 + x 2 d x {\displaystyle \pi ={\frac {25}{8}}+{\frac {1}{4}}\int _{0}^{1}{\frac {x(1-x)^{4}(1+4x+x^{2})}{1+x^{2}}}dx}
π = 22 7 − 1 28 ∫ 0 1 x ( 1 − x ) 8 ( 2 + 7 x + 2 x 2 ) 1 + x 2 d x {\displaystyle \pi ={\frac {22}{7}}-{\frac {1}{28}}\int _{0}^{1}{\frac {x(1-x)^{8}(2+7x+2x^{2})}{1+x^{2}}}dx}
π = 47 15 + ∫ 0 1 x 2 ( 1 − x ) 4 1 + x 2 d x {\displaystyle \pi ={\frac {47}{15}}+\int _{0}^{1}{\frac {x^{2}(1-x)^{4}}{1+x^{2}}}dx}
π = 47 15 + 2 ∫ 0 1 x 5 ( 1 − x ) 2 1 + x 2 d x {\displaystyle \pi ={\frac {47}{15}}+2\int _{0}^{1}{\frac {x^{5}(1-x)^{2}}{1+x^{2}}}dx}
π = 333 106 + 1 53 ∫ 0 1 x 4 ( 1 − x ) 4 ( 7 − 30 x + 60 x 2 − 30 x 3 ) 1 + x 2 d x {\displaystyle \pi ={\frac {333}{106}}+{\frac {1}{53}}\int _{0}^{1}{\frac {x^{4}(1-x)^{4}(7-30x+60x^{2}-30x^{3})}{1+x^{2}}}dx}
π = 377 120 − 1 2 ∫ 0 1 x 5 ( 1 − x ) 6 1 + x 2 d x {\displaystyle \pi ={\frac {377}{120}}-{\frac {1}{2}}\int _{0}^{1}{\frac {x^{5}(1-x)^{6}}{1+x^{2}}}dx}
π = 864 275 − 1 20 ∫ 0 1 x 4 ( 1 − x ) 6 ( 7 + 10 x + 7 x 2 ) 1 + x 2 d x {\displaystyle \pi ={\frac {864}{275}}-{\frac {1}{20}}\int _{0}^{1}{\frac {x^{4}(1-x)^{6}(7+10x+7x^{2})}{1+x^{2}}}dx}
π = 11 3 6 − 3 ∫ 0 1 x 2 ( 1 − x ) 2 1 + x + x 2 d x {\displaystyle \pi ={\frac {11{\sqrt {3}}}{6}}-{\sqrt {3}}\int _{0}^{1}{\frac {x^{2}(1-x)^{2}}{1+x+x^{2}}}dx}
π = 457 3 252 + 3 3 ∫ 0 1 x 4 ( 1 − x ) 4 1 + x + x 2 d x {\displaystyle \pi ={\frac {457{\sqrt {3}}}{252}}+{\frac {\sqrt {3}}{3}}\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x+x^{2}}}dx}
π = 5387 3 2970 − 3 9 ∫ 0 1 x 6 ( 1 − x ) 6 1 + x + x 2 d x {\displaystyle \pi ={\frac {5387{\sqrt {3}}}{2970}}-{\frac {\sqrt {3}}{9}}\int _{0}^{1}{\frac {x^{6}(1-x)^{6}}{1+x+x^{2}}}dx}
π = 17647759 3 9729720 + 3 27 ∫ 0 1 x 8 ( 1 − x ) 8 1 + x + x 2 d x {\displaystyle \pi ={\frac {17647759{\sqrt {3}}}{9729720}}+{\frac {\sqrt {3}}{27}}\int _{0}^{1}{\frac {x^{8}(1-x)^{8}}{1+x+x^{2}}}dx}
π = 9 3 5 + 3 ∫ 0 1 x 3 ( 1 − x ) 2 ( 1 + x ) 1 + x + x 2 d x {\displaystyle \pi ={\frac {9{\sqrt {3}}}{5}}+{\sqrt {3}}\int _{0}^{1}{\frac {x^{3}(1-x)^{2}(1+x)}{1+x+x^{2}}}dx}
π = 9 3 5 + 6 3 5 ∫ 0 1 x 3 ( 1 − x ) 2 1 + x 2 + x 4 d x {\displaystyle \pi ={\frac {9{\sqrt {3}}}{5}}+{\frac {6{\sqrt {3}}}{5}}\int _{0}^{1}{\frac {x^{3}(1-x)^{2}}{1+x^{2}+x^{4}}}dx}
π = 2 3 − 2 3 ∫ 0 1 x 4 1 + x 2 + x 4 d x {\displaystyle \pi =2{\sqrt {3}}-2{\sqrt {3}}\int _{0}^{1}{\frac {x^{4}}{1+x^{2}+x^{4}}}dx}
π = 789 3 435 + 30 3 145 ∫ 0 1 x 7 ( 1 − x ) 6 1 + x 2 + x 4 d x {\displaystyle \pi ={\frac {789{\sqrt {3}}}{435}}+{\frac {30{\sqrt {3}}}{145}}\int _{0}^{1}{\frac {x^{7}(1-x)^{6}}{1+x^{2}+x^{4}}}dx}
π = 1656 3 913 − 6 3 83 ∫ 0 1 x 6 ( 1 − x ) 8 1 + x 2 + x 4 d x {\displaystyle \pi ={\frac {1656{\sqrt {3}}}{913}}-{\frac {6{\sqrt {3}}}{83}}\int _{0}^{1}{\frac {x^{6}(1-x)^{8}}{1+x^{2}+x^{4}}}dx}
π = 8 ( 1 + 2 ) ∫ 0 1 x 2 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi =8\left(1+{\sqrt {2}}\right)\int _{0}^{1}{\frac {x^{2}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 8 ( 3 + 2 2 ) ∫ 0 1 x ( 1 − x ) 2 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi =8\left(3+2{\sqrt {2}}\right)\int _{0}^{1}{\frac {x(1-x)^{2}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 2 ( 4 + 3 2 ) ∫ 0 1 ( 1 − x ) 4 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi =2\left(4+3{\sqrt {2}}\right)\int _{0}^{1}{\frac {(1-x)^{4}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 8 ( 2 − 1 ) − 8 ( 2 − 1 ) ∫ 0 1 x 6 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi =8\left({\sqrt {2}}-1\right)-8\left({\sqrt {2}}-1\right)\int _{0}^{1}{\frac {x^{6}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 12 7 ( 2 2 − 1 ) + 8 7 ( 2 2 − 1 ) ∫ 0 1 x 5 ( 1 − x ) 2 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {12}{7}}\left(2{\sqrt {2}}-1\right)+{\frac {8}{7}}\left(2{\sqrt {2}}-1\right)\int _{0}^{1}{\frac {x^{5}(1-x)^{2}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 24 2 11 + 8 11 ∫ 0 1 x ( 1 − x ) 2 ( 1 + 2 2 x 4 ) 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {24{\sqrt {2}}}{11}}+{\frac {8}{11}}\int _{0}^{1}{\frac {x(1-x)^{2}(1+2{\sqrt {2}}x^{4})}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 20 2 9 − 2 2 3 ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {20{\sqrt {2}}}{9}}-{\frac {2{\sqrt {2}}}{3}}\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 35 191 ( 3 + 10 2 ) + 4 191 ( 3 + 10 2 ) ∫ 0 1 x 3 ( 1 − x ) 6 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {35}{191}}\left(3+10{\sqrt {2}}\right)+{\frac {4}{191}}\left(3+10{\sqrt {2}}\right)\int _{0}^{1}{\frac {x^{3}(1-x)^{6}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 126 ( 8 + 17 2 ) 1285 − 8 + 17 2 257 ∫ 0 1 x 2 ( 1 − x ) 8 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {126\left(8+17{\sqrt {2}}\right)}{1285}}-{\frac {8+17{\sqrt {2}}}{257}}\int _{0}^{1}{\frac {x^{2}(1-x)^{8}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 154 ( 33 + 58 2 ) 5639 + 2 ( 33 + 58 2 ) 5639 ∫ 0 1 x ( 1 − x ) 10 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {154\left(33+58{\sqrt {2}}\right)}{5639}}+{\frac {2\left(33+58{\sqrt {2}}\right)}{5639}}\int _{0}^{1}{\frac {x(1-x)^{10}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 858 ( 62 + 99 2 ) 55153 − 62 + 99 2 15758 ∫ 0 1 ( 1 − x ) 12 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {858\left(62+99{\sqrt {2}}\right)}{55153}}-{\frac {62+99{\sqrt {2}}}{15758}}\int _{0}^{1}{\frac {(1-x)^{12}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 173 ( 16 + 17 2 ) 2205 + 16 + 17 2 161 ∫ 0 1 x 6 ( 1 − x ) 8 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {173\left(16+17{\sqrt {2}}\right)}{2205}}+{\frac {16+17{\sqrt {2}}}{161}}\int _{0}^{1}{\frac {x^{6}(1-x)^{8}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 691 ( 11 + 10 2 ) 5530 − 4 ( 11 + 10 2 ) 79 ∫ 0 1 x 7 ( 1 − x ) 6 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {691\left(11+10{\sqrt {2}}\right)}{5530}}-{\frac {4\left(11+10{\sqrt {2}}\right)}{79}}\int _{0}^{1}{\frac {x^{7}(1-x)^{6}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 1838 ( 62 + 99 2 ) 118185 − 62 + 99 2 15758 ∫ 0 1 x 8 ( 1 − x ) 12 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {1838\left(62+99{\sqrt {2}}\right)}{118185}}-{\frac {62+99{\sqrt {2}}}{15758}}\int _{0}^{1}{\frac {x^{8}(1-x)^{12}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 10810 ( 8 + 17 2 ) 110253 − 8 + 17 2 257 ∫ 0 1 x 10 ( 1 − x ) 8 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {10810\left(8+17{\sqrt {2}}\right)}{110253}}-{\frac {8+17{\sqrt {2}}}{257}}\int _{0}^{1}{\frac {x^{10}(1-x)^{8}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 210798485 ( 424 + 577 2 ) 83203142022 + 424 + 577 2 972164 ∫ 0 1 x 10 ( 1 − x ) 16 1 + x 2 + x 4 + x 6 d x {\displaystyle \pi ={\frac {210798485\left(424+577{\sqrt {2}}\right)}{83203142022}}+{\frac {424+577{\sqrt {2}}}{972164}}\int _{0}^{1}{\frac {x^{10}(1-x)^{16}}{1+x^{2}+x^{4}+x^{6}}}dx}
π = 9 − e 2 + ∫ 0 1 ( 1 − x ) 2 4 ( e x − 1 − x − x 2 2 − 2 x 5 ( 1 − x ) 4 1 + x 2 ) d x {\displaystyle \pi ={\frac {9-e}{2}}+\int _{0}^{1}{\frac {(1-x)^{2}}{4}}\left(e^{x}-1-x-{\frac {x^{2}}{2}}-{\frac {2x^{5}(1-x)^{4}}{1+x^{2}}}\right)dx}