Page:Annales de mathématiques pures et appliquées, 1820-1821, Tome 11.djvu/114

106

INTÉGRATION

$V=x-x^{3},\quad X=4x,\quad X'=1+x^{2},\quad X''=0.$

De là nous conclurons successivement

${\frac {\operatorname {d} V}{\operatorname {d} x}}=1-3x^{2},\qquad X-X''=4x,$

${\begin{aligned}R\ =&{\frac {\operatorname {d} V}{\operatorname {d} x}}+(X-X'')=1+4x-3x^{2},\\R'=&{\frac {\operatorname {d} V}{\operatorname {d} x}}-(X-X'')=1-4x-3x^{2},\end{aligned}}$

${\begin{aligned}S\ =&\quad X+X'+X''=1+4x+x^{2},\\S'=&-X+X'-X''=1r-x+x^{2},\end{aligned}}$

${\begin{alignedat}{2}{\frac {\operatorname {d} R}{\operatorname {d} x}}=&+4-6x,\qquad &{\frac {\operatorname {d} R'}{\operatorname {d} x}}=&-4-6x,\\{\frac {\operatorname {d} S}{\operatorname {d} x}}=&+4+2x,&{\frac {\operatorname {d} S'}{\operatorname {d} x}}=&-4+2x\end{alignedat}}$

${\begin{aligned}S\ {\frac {\operatorname {d} V}{\operatorname {d} x}}=&1+4x-2x^{2}-12x^{3}-3x^{4},\\S'{\frac {\operatorname {d} V}{\operatorname {d} x}}=&1-4x-2x^{2}+12x^{3}-3x^{4},\\V\ {\frac {\operatorname {d} S}{\operatorname {d} x}}=&+4x+2x^{2}-4x^{3}-2x^{4},\\V'{\frac {\operatorname {d} S'}{\operatorname {d} x}}=&-4x+2x^{2}+4x^{3}-2x^{4},\\R\ {\frac {\operatorname {d} S}{\operatorname {d} x}}\ =&+4+18x-4x^{2}-6x^{3}.\end{aligned}}$